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The Best Mental Math Tricks

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Copyright Presh Talwalkar



About The Author

Presh Talwalkar studied Economics and Mathematics at Stanford University. His site Mind Your Decisions has blog posts and original videos about math that have been viewed millions of times.

Books By Presh Talwalkar

The Joy of Game Theory: An Introduction to Strategic Thinking. Game Theory is the study of interactive decision-making, situations where the choice of each person influences the outcome for the group. This book is an innovative approach to game theory that explains strategic games and shows how you can make better decisions by changing the game.

Math Puzzles Volume 1: Classic Riddles in Counting, Geometry, Probability, and Game Theory. This book contains 70 interesting brain-teasers.

Math Puzzles Volume 2: What do an Infinite Tower, a Classic Physics Puzzle, and Coin Flipping Have in Common? This is a follow-up puzzle book with 45 delightful puzzles.

But I only got the soup! This fun book discusses the mathematics of splitting the bill fairly.

40 Paradoxes in Logic, Probability, and Game Theory. Is it ever logically correct to ask “May I disturb you?” How can a football team be ranked 6th or worse in several polls, but end up as 5th overall when the polls are averaged? These are a few of the thought-provoking paradoxes covered in the book.

Multiply By Lines. It is possible to multiply large numbers simply by drawing lines and counting intersections. Some people call it “how the Japanese multiply” or “Chinese stick multiplication.” This book is a reference guide for how to do the method and why it works.



Table of Contents



Why Learn Mental Math Tricks?

Part I: Introductory Tricks

Calculate Percentages

The Rule Of 72

Estimate Your Hourly Wage

Calculate The Day Of The Week For A Date

The Times Table 11 To 20

Multiply By 7, Then 11, Then 13

Regroup Factors

Split Up A Multiplication

Divide By 4, 8, Etc.

Divide By 5, 50, Etc.



Part II: Squaring Numbers

Square A Number Ending In 5

Square A Number Ending In 9 Or 4

Square A Number ; Ending In 1 Or 6

Square A Number Ending In 8 Or 3, Or A Number Ending In 7 Or 2

Square A Two-Digit Number General Procedure

Square A Number Between 41 And 59

Square A Number Around 500

Square A Two-Digit Number Faster

Square 34, 334, Etc. Or Square 67, 667, Etc.



Part III: Multiplying Numbers

Multiply Two Numbers Ending In 5

Multiply Numbers Differing By 2

Multiply Numbers Units Digits Sum To 10 And Same Tens Digit

Multiply Numbers Tens Digits Sum To 10 And Same Units Digit

Multiply A Two-Digit Number By 11

Multiply Any Number By 11

Multiply Two Numbers In The 90s

Multiply Two Numbers From 101 To 109

Multiply Two Numbers Near 1,000

Multiply Two Numbers From 201 To 209, 301 To 309, Etc.

Multiply Two Numbers Just Below 200, 300, Etc.

Multiply Two Numbers Near 100 (One Below And One Above)

Multiply Two Numbers Near 200, 300, Etc. (One Below And One Above)



Part IV: Dividing Numbers

Divisibility Rules

Divide By 9

Find The Decimal Part When Dividing By 7

Find The Decimal Part When Dividing By 9, 99, Etc.

Find The Decimal Part When Dividing By 11, 101, Etc.

Find The Decimal Part When Dividing By 91

Find The Decimal Part When Dividing By 19

Find The Decimal Part When Dividing By 29, 39, Etc.

Find The Decimal Part When Dividing By 21

Find The Decimal Part When Dividing By 31, 41, Etc.



Part V: Concluding Tricks

Sum Odd Numbers

Sum Even Numbers

Sum Consecutive Numbers

Sum Fibonacci Style Sequences

Create a 3x3 Magic Square

Create a 4x4 Magic Square From Your Birthday

Convert a Decimal Number to Binary

The Egyptian Method / Russian Peasant Multiplication

Extract Cube Roots

Extract Fifth Roots

Extract Odd-Powered Roots



Conclusion

More From Presh Talwalkar





Why Learn Mental Math Tricks?

Mental math has a mixed reputation. Some consider it useless because calculators and computers can solve problems faster, with assured accuracy. Additionally, mental math is not even necessary to get good grades in math or to pursue a professional math career. So what's the point of learning mental math and math tricks anyway?

There are many reasons why mental math is still useful. For one, math skills are needed for regular tasks like calculating the tip in a restaurant or comparison shopping to find the best deal. Second, mental math tricks are one of the few times people enjoy talking about math. Third, mental math methods can help students build confidence with math and numbers.

Mental math tricks are fun to share. Imagine your friend asks you to multiply 93 and 97, and before they can type it into a calculator, you can figure out the answer in your head. Or imagine your friend asks you to divide 17 by 91, and you can calculate out the exact decimal answer in your head. Mental math tricks are impressive to see and they can be exciting to learn.

Mental math reduce the chore of simple math, build confidence in math skills, and serve as educational entertainment at parties and pubs. The flip side is some mental math calculations require extensive memorization and lots of practice to learn. This book is therefore not a comprehensive collection of mental math methods. This book seeks the middle ground: tricks that are relatively easy to learn, are fun, or have educational value.

Each section explains a math trick, and then has practice problems with complete solutions to illustrate the method. Every method is accompanied with a mathematical proof to justify why the trick works. While I find the proofs the most interesting parts, they can be skipped without detriment to learning the methods. The book can be read from start to finish. Or you can skip around, as most sections are independent of each other, excluding a few tricks that build upon previous ones.





Part I: Introductory Tricks


Calculate Percentages

While you can use a calculator to figure out percentages, there is a simple method you can use to solve them in your head. The best part is you probably are familiar with the basic skills for calculating percentages, which are calculating 10% and 50%.

Calculating 10%. Calculating 10 percent of a number is the same as dividing the number by 10. The procedure to divide a number by 10 is to move the decimal point one spot to the left. For example, to calculate 10% of 123 is the same as dividing 123 by 10. This is accomplished by moving the decimal point over 1 spot to the left, which means that 12.3 is 10% of 123. Similarly 10% of 56 is 5.6 and 10% of 4,987.56 is 498.756.

Calculating 50%. Calculating 50 percent of a number is the same as dividing the number by 2, which means to halve the number. For example, to calculate 50% of 123 is the same as halving 123. So 50% of 123 is 61.5. Similarly 50% of 56 is 28 and 50% of 5,000 is 2,500.

Calculating Other Percentages. You can calculate most percentages in your head if you can calculate 10% and 50%! The reason is that many percentages can be found as a combination of 10% and 50%.

Let's do an example of calculating 15% of 63. Note that 15% is equal to 10% plus 5%. To do this, you can calculate 10% and then you can add half of 10% (which is 5%) to itself. So we calculate 10% of 63 is equal to 6.3. Then we calculate 5% as half of that, which is 3.15. So we add these together to get 9.45, which is 15% of 63.

Let's do another example of 25% of 84. Note that 25% is equal to half of 50%. So we can calculate 25% of 84 by finding 50% and then halving the result. We can readily calculate that 50% of 84 is equal to 42, and then we need to take half of that number to get 21. Therefore we have 25% of 84 is equal to 21.

It is easy to calculate 10% and 50% in your head. From those starting points, you can easily find 5% (which is half of 10%), and then you can calculate all of the common percentages that are multiples of 5, 10, or 25.

Here is a short guide.

5% = half of 10%

10% = divide by 10

15% = add 10% and half of 10%

20% = find 10% and then double it

25% = find 50% and then halve it

30% = find 10% and then triple it

40% = find 10% and 50%. Subtract 10% value from 50% value

50% = divide in half

60% = find 10% and 50%. Add 10% value to 50% value

75% = find 50% and 25% and add together

80% = find 20% (as double 10%) and subtract from the whole

90% = find 10% and subtract from the whole

There can be more than one way to calculate a given percentage. For example, you can also calculate 60% by first finding 10% and then multiplying that value by 6.

The point is that you can calculate many common percentages by knowing how to calculate 10% and 50% in your head—and you probably can do these things without too much difficulty.

This trick can also help you estimate. For example, let's say you want to find 35% of 4,987. We can round 4,987 to a value of 5,000. Now we can calculate 10% is equal to 500, and 5% is half of that, or 250. So we can calculate 30% is triple 10% and equal to 1,500. We then add 5% and get to 1,750. This is very close to the exact answer of 1,745.45 for 35% of 4,987.

Now let's solve some practice problems.

Practice Problems

25% of 88

15% of 65

75% of 92

30% of 168

90% of 65

Proof

The method works because we can decompose a percentage like 15% into 15% = 10% + 5% = 10% + (10%)/2. In other words, it works because we can decompose many percentages into the operations of calculating 10% and calculating 50%.

Solutions to Practice Problems

25% of 88. We first calculate 50% by halving 88 to get 44. Then we divide that in half to find 25% is equal to 22.

15% of 65. We first calculate 10% which is 6.5, and then we divide that in half to find 5% is equal to 3.25. We add those results to get 15% of 65 is equal to 9.75.

75% of 92. We first calculate 50% by halving 92 to get 46. Then we divide that in half to find 25% is equal to 23. Subtracting 25% from the whole gets us that 75% of 92 is equal to 69.

30% of 168. We first calculate 10% which is 16.8, and we halve to get 5% is 8.4. Since 50% of 168 is 84, that means 25% is half of that, or 42. We add the 25% and 5% to get that 30% of 168 is 50.4.

90% of 65. We first calculate 10% by moving the decimal point to get 6.5. Then we subtract 10% from the whole to get 90% of 65 is equal to 58.5.



The Rule Of 72

You invest $1,000 at a guaranteed 4% interest rate annually. How long will it take your investment to double to $2,000?

We can solve the problem by computing compound interest. In the first year, you earn 4% on $1,000 which is $40. In other words, your investment grows to $1,000(1.04) = $1,040. In the next year, you earn another 4% on top of $1,040, which means your investment grows to $1,040(1.04) = $1,081.6 = $1,000(1.04)2. Noting the pattern, we can see that in T years your investment will grow to $1,000(1.04)T. We can set this equation equal to $2,000 and solve (using logarithms) that T is about 17.67 years. In other words, it takes about 18 years for your investment to double.

The Rule of 72 is a shortcut to estimate this answer very quickly. You divide 72 by the interest rate, expressed as a percentage, to get the doubling time.

For example, to calculate the doubling time with a fixed 4% interest rate, we divide 72 by the interest rate of 4 percent. Since 72/4 = 18, that means the investment will double in roughly 18 years. This is pretty close to the exact answer of 17.67!

Similarly, we can solve that an investment with 6% interest will double in roughly 72/6 = 12 years, an investment with 8% interest will double in roughly 72/8 = 9 years, and an investment with 12% interest will double in roughly 72/12 = 6 years. These are all very close to the exact answers, and the rule avoids having to compute compound interest.

There are two important caveats when using the rule of 72. First, the rule is a good approximation but it is not an exact answer. When you are working with investments, you may want to check an exact answer with a calculator or spreadsheet.

Second, the rule of 72 works only if the interest rate is fixed. When you invest in the stock market, the rate of return varies over time, and so you cannot use the rule of 72 with the same accuracy as you could for an investment with a fixed and guaranteed return.

Practice Problems

Calculate the doubling times for the following fixed interest rates.

2%

5%

7%

10%

20%

Proof

The rule works because it is an approximation to the compound interest formula. In order for an investment of R percent to double in T years, we need the interest multiplier (1 + R)T to be equal to 2. We can solve for T by using the natural logarithm to get T = ln(2)/ln(1 + R). We then use the approximations that ln(2) = 0.69 and ln(1 + R) = R (for values of R less than 30 percent). Therefore, we have T = 0.69/R.

Now we multiply by 100/100 to get rid of the decimal points. Thus, we have T = 69/(100R), so we can consider the interest rate R as a whole number instead of a percentage (4% is just 4). Finally we get a slightly better approximation, and have more factors to divide by evenly, when we make the numerator 72 instead of 69. So we have T = 72/(100R).

Solutions to Practice Problems

2% will double in about 72/2 = 36 years

5% will double in about 72/5 = 14.4 years

7% will double in about 72/7 = 10.3 years

10% will double in about 72/10 = 7.2 years

20% will double in about 72/20 = 3.6 years



Estimate Your Hourly Wage

Let's say you earn $50,000 in a full-time job. What is your hourly wage? A good estimate is to divide your compensation by 2,000. This can be done expediently by halving your compensation and then dividing by 1,000 (moving the decimal 3 spots to the left).

So let's divide $50,000 by 2,000. We halve $50,000 to get $25,000, and then we divide by 1,000 to get $25 per hour.

Practice Problems

Calculate the hourly wage for the following annual incomes.

$20,000

$60,000

$100,000

$250,000

$1,000,000

Proof

In America, a full-time job typically means 40 hours per week for about 50 weeks, which translates into 40 x 50 = 2,000 hours in a year. So you can find your hourly wage by dividing your annual salary by 2,000.

Solutions to Practice Problems

$20,000 is about an hourly wage of $20,000/2,000 = $10

$60,000 is about an hourly wage of $60,000/2,000 = $30

$100,000 is about an hourly wage of $100,000/2,000 = $50

$250,000 is about an hourly wage of $250,000/2,000 = $125

$1,000,000 is about an hourly wage of $1,000,000/2,000 = $500



Calculate The Day Of The Week For A Date

January 1, 2015 fell on a Thursday. Which day of the week corresponds to the date of December 25, 2015?

There are many ways to find the answer. Perhaps the simplest is to look up the answer in a calendar, a spreadsheet, or a website like WolframAlpha. But it is actually not too difficult to figure this out in your head, with a little bit of memorization and some practice.

The reason is the calendar days follow a regular pattern. Every 7 days cycles to the same day of the week. Because January 1, 2015 was a Thursday, that means the dates 1/8, 1/15, 1/22, and 1/29 must also fall on a Thursday. Similarly, the dates 1/3, 1/10, 1/17, 1/24, and 1/31 must be on a Saturday, because those dates are two days following the known date of Thursday January 1.

The mathematician John Conway devised a method called the Doomsday Rule that uses the pattern to allow easy mental calculation the day of the week for any date in a year. The method starts by memorizing dates in each month that fall on the same day of the week, known as a doomsday.

One set of dates are the doubles. In a year, the dates 4/4, 6/6, 8/8, 10/10, and 12/12 always fall on the same day of the week. The next set of dates falling on the same day of the week is 5/9, 7/11, 9/5, and 11/7. This can be memorized by the phrase “I work 9 to 5 at the store 7-11.” We only need reference dates for the first three months now. In March, the date 3/14 is a doomsday (this is Pi Day because the constant π has the decimal expansion 3.14...). The last date in February is also on the same day, either 2/28 in a non-leap year or 2/29 in a leap year. Finally, the relevant date in January is 1/3 in a non-leap year and 1/4 in a leap year.

If you know the doomsday for a year, then you can find the day of the week for the remaining dates from the reference dates. Let's do some examples. We know January 3, 2015 falls on a Saturday. Which day does September 7, 2015 fall on? The closest reference date is 9/5, which also falls on a Saturday. Thus, September 7 is two days later on a Monday.

Now let's figure out Christmas December 25, 2015. The closest special date is 12/12, which falls on a Saturday. The 25th can be found by computing the difference in the number of days from the known date. That is, we calculate 25 – 12 = 13 days, which means 1 week, 6 days. Therefore, Christmas is one day before two weeks from Saturday, so it must be on a Friday.

The trick is to use the special dates as anchors and then figure out any other date relative to that. For practical purposes, you are almost always dealing with dates in the current year, so you only have to memorize the doomsday's day of the week for the current year.

But John Conway was not content with only the current year. He figured out how to generalize the rule to figure out the date in any year! This requires just a bit more memorization and calculation, but it allows you to figure out the day of the week for practically any date.

It will be useful to think about the days of the week in numerical codes. Here is a code with mnemonic tips, developed by Conway.

0 = Sunday = Nones-day

1 = Monday = One-day

2 = Tuesday = Twos-day

3 = Wednesday = Trebles-day

4 = Thursday = Fours-day

5 = Friday = Five-day

6 = Saturday = Six-a-day

We then need to memorize when the doomsday code for years ending with 00. These will serve as anchor days in our calculations.

1800 – 1899 = Friday = day code 5

1900 – 1999 = Wednesday = day code 3

2000 – 2099 = Tuesday = day code 2

2100 – 2199 = Sunday = day code 0

Now we can find the doomsday for any year from 1800 to 2199. Let's write the date as CCYY where CC is the century and YY is the last two digits. Here is the formula for the doomsday in a given year.

Doomsday code = [YY + rounddown(YY/4) + anchor(CC)] mod 7

Let me explain the terms in this formula. The term rounddown(YY/4) means to take the year YY, divide it by 4, and then round-down the result. For example, in 2015, we would have YY = 15 and so YY/4 = 15/4 = 3.75. We need to round-down the result, so the term is 3.

The term anchor(CC) means to take the anchor day code for the given century. In 2015, the century is the 2000s, and that anchor day is a Tuesday, with a day code of 2.

The final part is to take the entire result modulo 7. That means to only consider the remainder after 7. Equivalently, that means to reduce the number by multiples of 7 until the result is between 0 and 6. The reason we do this is we want to get a day code between 0 and 6, and after all, every 7 days corresponds to the same day of the week.

So let's use the formula on the year 2015. We will have:

Doomsday code = [YY + rounddown(YY/4) + anchor(CC)] mod 7

Doomsday code = [15 + rounddown(15/4) + anchor(20)] mod 7

Doomsday code = [15 + 3 + 2] mod 7

Doomsday code = [20] mod 7

Doomsday code = 6

So the doomsday in 2015 is a 6, or a Saturday. The key in knowing the doomsday for 2015 is we know all of the special dates fall on a Saturday as well, and other dates can be determined in relation to them.

Now let's put everything we've learned to solve a couple of examples. Let's find the day of the week for January 1, 1903. We first find the doomsday code, remembering the anchor for 1900s is a Wednesday (day code 3).

Doomsday code = [03 + rounddown(03/4) + anchor(19)] mod 7

Doomsday code = [3 + 0 + 3] mod 7

Doomsday code = 6

Therefore the Doomsday in 1903 was a Saturday. Since 1903 was not a leap year (leap years are divisible by 4), the closest memorable date is 1/3 which would also be a Saturday. Therefore January 1, 1903 fell two days earlier on a Thursday.

Let's do one more example of February 4, 2116. We first find the doomsday code, remembering the anchor for 2100s is a Sunday (day code 0).

Doomsday code = [16 + rounddown(16/4) + anchor(21)] mod 7

Doomsday code = [16 + 4 + 0] mod 7

Doomsday code = 20 mod 7

Doomsday code = 6

Therefore the Doomsday in 2116 falls on a Saturday. Since 2116 is a leap year (leap years are divisible by 4), the closest memorable date is the last day of February, 2/29, which would also be a Saturday. Then 2/4 would be 25 days earlier, which is 3 weeks and 4 days earlier. So four days before Saturday is a Tuesday, and therefore February 4, 2116 will be on a Tuesday.

I will mention there is also a special rule for century leap years. A year ending in 00 is only a leap year if it is also divisible by 400. So 2000 is a leap year, but 1800, 1900, and 2100 are not leap years.

Practice Problems

March 15, 2015

January 1, 2000

July 4, 1876

December 25, 2150

July 16, 1969

Proof

There are two parts we will justify. First, let's consider a specific year and the memorable dates. We claimed the same day of the week happens for the dates 4/4, 6/6, 8/8, 10/10, 12/12, 5/9, 9/5, 7/11, 11/7, and 3/14. This is a consequence that the same day of the week happens every 7 days, and each month has a specified number of dates. It just so happens that these dates are all multiples of 7 days apart, and hence they fall on the same day of the week. The dates for January and February depend on whether it is a leap year, because the extra day for the leap year changes which dates are multiples of 7 days apart. In a non-leap year, there is no extra day and the dates are 1/3 and 2/28. In a leap year, the extra day means the dates are 1/4 and 2/29.

The next part to justify is the formula for calculating the day of the week the doomsday falls on during a given year. Here is the formula.

Doomsday code = [YY + rounddown(YY/4) + anchor(CC)] mod 7

Why does this formula work? The formula starts with the doomsday for the 00 year of a given century, which is denoted anchor(CC). For any other year, YY, in that same century, we need to count the number of days that have passed. A normal year adds 365 days and a leap year adds 366 days. Since every 7 days is the same day of the week, we can reduce these numbers modulo 7 to find the offset. So a normal year adds 1 day, and a leap year adds 1 more day. So we can find the doomsday for the remaining years in a century by counting the number of years that have passed from 00, which is YY, and then adding in an extra day for each leap year, which is rounddown(YY/4). So this formula exactly accounts for the offset of each new year and leap year, and therefore it gives the doomsday for a year YY in a century CC.

Solutions to Practice Problems

March 15, 2015. The century code for 2000s is 2. So the doomsday in 2015 is calculated as (15 + rounddown(15/4) + 2) mod 7, which is then equal to 20 mod 7 = 6. So the doomsday falls on day code 6, a Saturday, and hence March 14 does as well. This means March 15, 2015 falls one day later on a Sunday.

January 1, 2000. The century code for 2000s is 2. So the doomsday in 2000 is calculated as (00 + rounddown(00/4) + 2) mod 7, which is then equal to 2 mod 7 = 2. So the doomsday falls on day code 2, a Tuesday, and hence in the leap year 2000 the date January 4 does as well. This means January 1, 2000 corresponds to 3 days earlier, on a Saturday.

July 4, 1876. The century code for 1800s is 5. So the doomsday in 1876 is calculated as (76 + rounddown(76/4) + 5) mod 7, which is then equal to 16 mod 7 = 2. So the doomsday falls on day code 2, a Tuesday, and hence July 11 (7/11) does as well. This means July 4, 1876 corresponds to one week earlier, also a Tuesday.

December 25, 2150. The century code for 2100s is 0. So the doomsday in 2150 is calculated as (50 + rounddown(50/4) + 0) mod 7, which is then equal to 13 mod 7 = 6. So the doomsday falls on day code 6, a Saturday, and hence December 12 (12/12) does as well. This means December 25, 2150 falls one week and six days later, on a Friday.

July 16, 1969. The century code for 1900s is 3. So the doomsday in 1969 is calculated as (69 + rounddown(69/4) + 3) mod 7, which is then equal to 89 mod 7 = 5. So the doomsday falls on day code 5, a Friday, and hence July 11 does as well. This means July 16, 2015 falls five days later on a Wednesday.



The Times Table 11 To 20

Can you multiply 12 by 14 easily? Problems up to 10 are often memorized when we learn the times table up to 10. This trick will help you figure out the times table from 11 to 20 by extending the times table up to 10.

There are four main steps. First, add the units digit of the second number to the first number. Second, multiply the result by 10. Third, multiply the units digits of both numbers. And finally, add the results from the second and third steps.

Let's do an example of 12 x 14. The number 14 has the units digit of 4. We add the units digit 4 to the number 12 to get 16. Then we multiply that by 10 to get 160. Third, we multiply the units digits of 2 and 4 to get 8. Finally, we add 160 and 8 to get 168.

Let's do another example of 15 by 18. We add 8 to 15 to get 23, and then we multiply that by 10 to get 230. We then multiply the units digits of 5 and 8 to get 40. Adding that to 230 gets an answer of 270.

Notice we would get the same answer even if we did 18 times 15. In this problem, we would add 5 to 18 to get 23, which we multiply by 10 to get 230. Then we multiply the units digits of 8 and 5 to get 40. Adding 230 and 40 once again results in the answer of 270.

This trick makes multiplying numbers between 11 and 20 an extension of the times tables from 1 to 10.

Practice Problems

11 x 16

12 x 17

14 x 19

15 x 13

16 x 16

Proof

Numbers between 11 and 19 can be written as 10 + x and 10 + y, for x and y between 1 and 9. The product of two numbers can then be found as (10 + x)(10 + y) = 100 + 10(x + y) + xy = 10[10 + (x + y)] + xy.

The term 10 + x + y means to add the units digit of one number to the other number. Then that is multiplied by 10, after which the product of the units digits, xy, is added. This is the procedure described in this trick.

Solutions to Practice Problems

11 x 16. We add 6 to 11 to get 17, which is multiplied by 10 to get 170. Then the term 1 x 6 = 6 is added to get 176.

12 x 17. We add 7 to 12 to get 19, which is multiplied by 10 to get 190. Then the term 2 x 7 = 14 is added to get 204.

14 x 19. We add 9 to 14 to get 23, which is multiplied by 10 to get 230. Then the term 4 x 9 = 36 is added to get 266.

15 x 13. We add 3 to 15 to get 18, which is multiplied by 10 to get 180. Then the term 5 x 3 = 15 is added to get 195.

16 x 16. We add 6 to 16 to get 22, which is multiplied by 10 to get 220. Then the term 6 x 6 = 36 is added to get 256.



Multiply By 7, Then 11, Then 13

Ask a friend to pick a three-digit number. Tell them to multiply the number by 7, then 11, and then 13 on a calculator. Before they have the result, you will already know the answer.

Let's see how this works. Imagine your friend picks 123. You can instantly say the final result is 123,123. Or if your friend picks 479, you know the answer is 479,479.

The pattern is very simple so you can probably only do this trick once. The trick is that a number ABC multiplied by 7, then 11, then 13 will be ABC, ABC as a six-digit number.

Practice Problems

378 x 7 x 11 x 13

926 x 7 x 11 x 13

Proof

Note that 7 x 11 x 13 = 1,001. When you multiply ABC by 1,001, that is the same as multiplying by (1,000 + 1). Multiplying by 1,000 shifts the number ABC by three spots to get ABC,000. Then we multiply by 1 and add the result, which means adding ABC to the number again. The final result is ABC, ABC as a six-digit number.

Solutions to Practice Problems

378 x 7 x 11 x 13. The answer is the three-digit number 378 repeated twice. The result is 378,378.

926 x 7 x 11 x 13. The answer is the three-digit number 926 repeated twice. The result is 926,926.



Regroup Factors

Can you multiply 25 by 80? Sometimes we can re-group factors of the numbers to make the problem simpler. For example, we can think about 80 as 4 x 20, so we have 25 x 80 = 25 x (4 x 20). We can then write this as (25 x 4) x 20 = 100 x 20 = 2,000. So we can also conclude 25 by 80 is equal to 2,000.

The trick of regrouping factors is about shifting around factors from both numbers to get more convenient numbers. Which numbers are “convenient?”

Mostly we look for ways to create factors of 10, because it is easy to multiply by 10. Here are some more problems we can use the method of regrouping factors.

Multiply 5 by 46. The trick is that 5 x 2 = 10, and since 46 is an even number it has a factor of 2. Therefore 5 x 46 = 5 x (2 x 23), and we can write this as (5 x 2) x 23 = 10 x 23. This last problem is easy as multiplying by 10 is the same as “adding a zero” to the end of the number. Thus, 5 x 46 = 230.

Multiply 15 by 66. We again want to find a factor of 10. We can factor a 5 from the number 15, and since 66 is even we can factor a 2 from it. Therefore, we have 15 x 66 = (3 x 5) x 66 = (3 x 5) x (2 x 33). Now we write 3 x (5 x 2) x 33 = 3 x 10 x 33 = (3 x 33) x 10 = 99 x 10 = 990. This might seem like a daunting sequence of steps, but it will become natural after you start looking for ways to regroup factors.

Regrouping factors is a method to create an equivalent multiplication problem that is easier to calculate. This mostly comes in handy when you have numbers that end in 5 multiplied by even numbers, which have a factor of 2. You can combine the factors of 5 and 2 to get a factor of 10, which makes the problem easier.

Or you might have multiple factors of 5 and 2. For example, let's do 125 times 44. The number 125 is 5 x 25 and the number 44 is 4 x 11. So we can calculate 125(44) = (5 x 25)(4 x 11) = 5(25 x 4)11, and then we finish the steps as 5(100)11 = 500(11) = 5,500.

Regrouping factors is a good mental math trick if you can pull it off. But it is also useful when you are doing problems by hand.

Practice Problems

25 x 16

75 x 6

150 x 80

35 x 12

45 x 8

Proof

Re-grouping works because multiplication is associative and commutative. That is, we can multiply numbers, or factors of numbers, in a re-grouped order and get the same result.

Solutions to Practice Problems

25 x 16. We re-group this problem as 25 x 16 = 25 x (4 x 4), which is then equal to (25 x 4) x 4 = 100 x 4 = 400.

75 x 6. We re-group this problem as 75 x 6 = 75 x (2 x 3), which is then equal to (75 x 2) x 3 = 150 x 3 = 450.

150 x 80. We re-group this problem as 150 x 80 = 150 x (4 x 20), which is then equal to (150 x 4) x 20 = 600 x 20 = 12,000.

35 x 12. We re-group this problem as 35 x 12 = 35 x (2 x 6), which is then equal to (35 x 2) x 6 = 70 x 6 = 420.

45 x 8. We re-group this problem as 45 x 8 = 45 x (2 x 4), which is then equal to (45 x 2) x 4 = 90 x 4 = 360.



Split Up A Multiplication

What is 93 times 7? This problem is a lot easier to solve in your head if you split it up into two easier problems. We can think about 93 = 90 + 3, and then we can do (90 + 3) x 7 = 90 x 7 + 3 x 7. While this might look more difficult, it is actually easier to do because the problem involves adding up the result of two easier tasks. Since we have 90 x 7 = 630 and 3 x 7 = 21, the answer is the sum of 651.

Let's try another problem. What is 93 x 17? Multiplying two-digit numbers in your head would be very hard to do the traditional way, as you have to remember to carry over. So we can split up the problem as the problem 93 x 17 = 93 x (20 – 3) = 93 x 20 – 93 x 3. The first part of the expression 93 x 20 = 1,860 is easy. But then to do 93 x 3, we again split it up as (90 + 3) x 3 = 90 x 3 + 3 x 3 = 270 + 9 = 279. So we subtract 279 from 1,860 to get 1,581. (Note the subtraction could be split up as well to 1,860 – 279 = 1,860 – 300 + 21 = 1,560 + 21 = 1,581.)

In general, try to re-write one of the numbers as a difference from a multiple of 10, 100, etc. and then you can break the problem down into easier parts.

Practice Problems

97 x 3

101 x 12

11 x 26

198 x 21

17 x 212

Proof

Splitting up a multiplication works because of the distributive property of multiplication, which states (x + y) z = xz + yz. So we can break up a number into two parts x and y and then add up the product of each part with z.

Solutions to Practice Problems

97 x 3. This problem can be split up as 97 x 3 = (100 – 3) x 3, which is then equal to 100(3) – (3)(3) = 300 – 9 = 291.

101 x 12. This problem can be split up as 101 x 12 = (100 + 1) x 12, which is then equal to 100(12) + (1)(12) = 1,200 + 12 = 1,212.

11 x 26. This problem can be split up as 11 x 26 = (10 + 1) x 26, which is then equal to 10(26) + (1)(26) = 260 + 26 = 286.

198 x 21. This problem can be split up as 198 x 21 = (200 – 2) x 21, which is then equal to 200(21) – (2)(21) = 4,200 – 42 = 4,158.

17 x 212. This problem can be split up as 17 x 212 = 17 x (200 + 12), which is then equal to 17(200) + 17(12) = 3,400 + 17(10 + 2). We can continue evaluating 3,400 + 17(10) + 17(2) = 3,400 + 170 + 34, and finally we have 3,570 + 34 = 3,604.



Divide By 4, 8, Etc.

What is 256 divided by 4? Instead of working it out by long division, you can divide by 4 by halving two times. So 256 halved is 128, which we then halve again to get 64.

Similarly, you can divide by 8 by halving three times. For example, let's do 1,024 divided by 8. We can halve 1,024 to get 512, then we halve that to get 256, and we halve one more time to get 128.

In general, you can divide by 2n if you repeatedly halve a total of n times.

Practice Problems

72 ÷ 4

135 ÷ 4

100 ÷ 8

1,234 ÷ 8

2,082 ÷ 16

Proof

Dividing a number x by 4 is equivalent to dividing x by 22. So we have x/4 = x/(22) = (x/2)(1/2) = (x/2)/2. In other words, we can divide by 4 by dividing by 2 twice.

Similarly, dividing a number x by 2n can be expressed as dividing by 2 a total of n times. So we have x/(2n) = (x/2)(1/2)(1/2)...(1/2), which is equal to (x/2)/2 … / 2, where we are dividing by 2 a total of n times.

Solutions to Practice Problems

72 ÷ 4. We need to halve two times to divide by 4 = 22. We halve 72 to get 36, which is then halved again to get 18.

135 ÷ 4. We need to halve two times to divide by 4 = 22. We halve 135 to get 67.5, which is then halved again to get 33.75.

100 ÷ 8. We need to halve three times to divide by 8 = 23. We halve 100 to get 50, then halve that to get 25, and finally halve that to get 12.5.

1,234 ÷ 8. We need to halve three times to divide by 8 = 23. We halve 1,234 to get 617, then we have again to get 308.5, and we halve a third time to get 154.25.

2,082 ÷ 16. We need to halve four times to divide by 16 = 24. We halve 2,082 to get 1,041, then we have again to get 520.5, then we halve a third time to get 260.25, and then we halve a fourth time to get 130.125.



Divide By 5, 50, Etc.

What is 350 divided by 5?

A simple way to find the answer is to note that 1/5 = 2/10. So dividing by 5 is the same as dividing by 10 and then multiplying by 2. So 350 divided by 10 is 35, which we then double to get 70.

This trick is useful for any number that starts with a 5 and has only 0s after that. For example, let's divide 1,235 by 50. We first divide 1,235 by 100 to get 12.35. Then we double that to get 24.70.

In general, to divide by 50..0, which has N zeros, you can divide by 100..0, which has N + 1 zeros, and then double the result. Alternately, you can double the result and then divide by 100..0, which has N + 1 zeros.

You can use a similar trick to divide by 25, 250, etc. The key is to observe that 1/25 = 4/100. So you can divide by 25 by dividing by 100 and then doubling two times. So what is 1,235 divided by 25? We first divide by 100 to get 12.35. Then we double to get 24.70, and then we double again to get 49.40.

In general, to divide by 250..0, which has N zeros, you can divide by 100..0, which has N + 2 zeros, and then quadruple the result by doubling two times. Alternately, you can do the quadrupling first and then divide by 100..0.

Practice Problems

315 ÷ 50

1,210 ÷ 50

1,324 ÷ 50

11,434 ÷ 500

22,082 ÷ 5,000

Proof

Since 50 = 100/2, dividing x by 50 means x/50 = x/(100/2) = (2x)/100. In other words, we can divide by 50 by doubling the number and then dividing by 100. Equivalently, we can divide by 100 and then double the number.

Similarly, dividing by 5(10n) is the same as doubling the number and then dividing by 10n+1. Therefore x/[5(10n)] = x/(10n+1/2) = (2x)/10n+1.

Dividing a number x by 25(10n) can be expressed as quadrupling the number and then dividing by 10n+2. The proof is x/[25(10n)] can be written as x/(10n+2/4) = (4x)/10n+2.

Solutions to Practice Problems

315 ÷ 50. We divide by 100 to get 3.15 and then double that to get 6.3.

1,210 ÷ 50. We divide by 100 to get 12.1 and then double to get 24.2.

1,324 ÷ 50. We divide by 100 to get 13.24 and then double to get 26.48.

11,434 ÷ 500. We divide by 1,000 to get 11.434 and then double to get 22.868.

22,082 ÷ 5,000. We divide by 10,000 to get 22.082 and then double to get 4.4164.





Part II: Squaring Numbers


Square A Number Ending In 5

Can you square 35 in your head? What about 65 squared? It is very easy to find the square of a number ending in 5.

There are two steps. First, multiply the tens digit by one more than itself. Second, append the number 25 to the result from step 1.

For example, let's calculate 352. The tens digit of 35 is 3. So we multiply 3 by one more than itself, which is 4, to get 3 x 4 = 12. The second step is to append 25 to the 12, which means the answer is 1,225.

For another example, let's do 652. The tens digit is 6 and one more than that is 7. So we multiply 6 by 7 to get 42. We then append 25 to the end of the number to get an answer of 4,225.

This trick actually works for three-digit numbers and larger numbers too. The only modification is the “tens” digit is the number resulting from deleting the 5 from the number. So in the number 125, make the “tens” digit the number by deleting the 5, which is the number 12.

For example, let's say we want to square 105. For the first step, we delete the 5 and we are left with the number 10. We multiply 10 by one more than itself, 11, to get 110. Then we append 25 to get 11,025 as the answer.

Take some time to learn and practice this rule because some of the later tricks in the book will build upon this trick.

Practice Problems

152

452

752

1252

2552

Proof

Squaring a number ending in 5 means to multiply (10x + 5) times itself. By algebra, we have (10x + 5)(10x + 5) = 100x2 + 100x + 25, which is then equal to 100x(x + 1) + 25. The term 25 gives the last two digits of 25, and the term 100x(x + 1) indicates to multiply the number x times one more than itself, x + 1, and then shift over two places so these are the leading digits of the answer.

Solutions to Practice Problems

152. We find the answer starts as 1 x 2 = 2 and then ends with 25. So the answer is 152 = 225.

452. We find the answer starts as 4 x 5 = 20 and then ends with 25. So the answer is 452 = 2,025.

752. We find the answer starts as 7 x 8 = 56 and then ends with 25. So the answer is 752 = 5,625.

1252. We find the answer starts as 12 x 13. There are many ways to solve this. We will re-write 12 x 13 = 12 x (12 + 1). Then we can solve the problem as 122 + 12 = 144 + 12 = 156. Those are the first three digits of the answer, and we append the digits 25. So the answer is 1252 = 15,625.

2552. We find the answer starts as 25 x 26 = 252 + 25. We can find 252 by the same trick. So we do 2 x 3 = 6 and append 25 to get 625. Thus we have 25 x 26 = 252 + 25 = 625 + 25 = 650. The answer then ends with the digits 25. So the answer is 2552 = 65,025.



Square A Number Ending In 9 Or 4

It is easy to square a number ending in 0 because a number ending in 0 is a multiple of 10. The rule is to square the leading digits and then double the number of 0s. For example, 302 can be calculated by squaring 3, which is 9, and then doubling the number of zeros to 00 to get the answer 900. Similarly, 802 is equal to 82 = 64 followed by 00, or 6,400.

It is also relatively easy to square a number ending in 5, as already explained. The rule is to multiply the leading digits by a number one more than itself and then append 25 to the result. For example, 352 can be calculated by multiplying 3 by 4, which is 12, and then appending 25 to get the answer 1,225. Similarly, 852 is equal to 8 x 9 = 72 followed by 25, or 7,225.

Numbers ending in 9 are one less than numbers ending in 0. Similarly, numbers ending in 4 are one less than numbers ending in 5. As such, we can square numbers ending in 9 or 4 by squaring numbers one more than them and then adjusting the result.

Here is the rule. Let's say the number ends in a 9 or 4. We can calculate by squaring one more than the number, and then we subtract the original number and one more than the original number.

Let's do some examples. Let's calculate 392. One more than the number is 40, and 402 = 1,600. We then subtract the original number, 39, and one more than the original number, 40. So we take 1,600 and subtract 39 and 40. The end result is 392 = 1,521 = 1,600 – 39 – 40.

Let's calculate 742. One more than the number is 75, and 752 = 5,625 by the rule of squaring a number ending in 5. We then subtract the original number, 74, and one more than the original number, 75. So we take 5,625 and subtract 74 and 75. The end result is 742 = 5,476.

To summarize, you can square a number ending in 9 or 4 by squaring the number one above it, and then subtracting out the original number and one more than the original number. You will find it easier to square the number one more—as it will be a number ending in 0 or 5—and then you can adjust the result by subtracting the original number and one more than the original number.

Practice Problems

142

492

842

1292

2542

Proof

Here is why the rule works. For any number x, you can obtain x2 by squaring one more than x and then subtracting x and x + 1. The algebraic proof is (x + 1)2 – x – (x + 1)= (x2 + 2x + 1) – 2x – 1 = x2.

Solutions to Practice Problems

142. We need to calculate 152 first. This is found by the rule of squaring a number ending in 5. So we calculate 1 x 2 = 2 and then append the digits 25, so 152 = 225. Then we have 142 = 152 – 14 – 15 = 225 – 29 = 196.

492. We calculate 502 = 2,500 first. Then we subtract 49 and 50 to get the result 492 = 2,401.

842. We need to calculate 852 first. This is found by the rule of squaring a number ending in 5. So we calculate 8 x 9 = 72 and then append the digits 25, so 852 = 7,225. Then we have 842 = 852 – 84 – 85 = 7,056.

1292. We calculate 1302 = 16,900 first. Then we subtract 130 and 129 to get 1292 = 16,900 – 260 + 1 = 16,641.

2542. We need to calculate 2552 first. This is found by the rule of squaring a number ending in 5. So we calculate 25 x 26 = 252 + 25. We can find 252 by the same trick. So we do 2 x 3 = 6 and append 25 to get 625. Thus 25 x 26 = 252 + 25 = 625 + 25 = 650. The answer then ends with 25, so 2552 = 65,025. Then we have 2542 = 2552 – 255 – 254, which is equal to 65,025 – 510 + 1 = 64,516.



Square A Number Ending In 1 Or 6

Numbers ending in 1 are one more than numbers ending in 0. Similarly, numbers ending in 6 are one more than numbers ending in 5. The rule for numbers ending in 1 or 6 is very similar to the one for numbers ending in 9 or 4.

Here is the rule. Let's say the number ends in a 1 or 6. We calculate by squaring one less than the number, and then we add back the original number and one less than the original number.

Let's do some examples. Let's calculate 212. One less than the number is 20, and 202 = 400. We then add the original number, 21, and one less than the original number, 20, for an adjustment of 41. The end result is 212 = 441.

Let's calculate 862. One less than the number is 85, and 852 = 7,225 (by the rule of squaring a number ending in 5). We then add the original number, 86, and one less than the original number 85, for an adjustment of 171. The end result is 862 = 7,396.

How can you remember the rule for a number ending in 4 or 9 versus the rule for a number ending in 1 or 6? The key is to think about how the original number compares to the convenient number. When the original number is smaller (like in 4 or 9), you need to adjust the square by subtracting the convenient number and the original number. When the original number is larger (like in 1 or 6), you need to adjust by adding the convenient number and the original number.

Practice Problems

162

412

762

1212

2562

Proof

Here is why the rule works. For any number x, you can obtain x2 by squaring one less than x and then adding x and x – 1. The algebraic proof is (x – 1)2 + x + (x – 1)= (x2 – 2x + 1) + (2x – 1) = x2.

Solutions to Practice Problems

162. We need to calculate 152 first. This is found by the rule of squaring a number ending in 5. So we calculate 1 x 2 = 2 and then append the digits 25, so 152 = 225. Then we have 162 = 152 + 15 + 16 = 225 + 31 = 256.

412. We calculate 402 = 1,600 first. Then we add 40 and 41 to get the answer 412 = 1,681.

762. We need to calculate 752 first. This is found by the rule of squaring a number ending in 5. So we calculate 7 x 8 = 56 and then append the digits 25, so 752 = 5,625. Then we have 742 = 752 + 75 + 76, which then equals 5,625 + 151 = 5,776.

1212. We calculate 1202 = 14,400 first. Then we add 120 and 121 to get 1212 = 14,400 + 241 = 14,641.

2562. We need to calculate 2552 first. This is found by the rule of squaring a number ending in 5. So we calculate 25 x 26 = 252 + 25. We can find 252 by the same trick. So we do 2 x 3 = 6 and append 25 to get 625. Thus 25 x 26 = 252 + 25 = 625 + 25 = 650. The answer then ends with 25, so 2552 = 65,025. Then we have 2562 = 2552 + 255 + 256, which is the result of 65,025 + 511 = 65,536.



Square A Number Ending In 8 Or 3, Or A Number Ending In 7 Or 2

For completeness, we will present similar rules for numbers ending in 8 or 3, and 7 or 2. These rules are not as easy to remember, but they are shown to illustrate how you can extend the previous rules for the remaining digits.

Here is the rule for 8 or 3. The convenient number is 2 more than the original number. We can calculate the square by calculating the square of the convenient number, subtracting 4 times the convenient number, and then adding back 4.

Let's do some examples. Let's calculate 382. The convenient number is 2 more than the original number, 40. So we have 402 = 1,600. We then quadruple the convenient number 40 to get 160. Subtracting 160 from 1,600 results in 1,440. Finally, we add 4 to get 1,444 = 382.

Let's calculate 732. Two more than the number is the convenient number of 75, and 752 = 5,625 by the rule of squaring a number ending in 5. We then quadruple 75 to get 300. Subtracting 300 from 5,625 results in 5,325. Finally, adding 4 means 732 = 5,329.

The rule for 7 or 2 is similar, but the nearby convenient number is two less than the number ending in 7 or 2. So we can calculate the square by calculating the square of the convenient number, adding 4 times the convenient number, and then adding back 4.

Let's do some examples. Let's calculate 622. The convenient number is two less than 62, which is 60, and 602 = 3,600. We then quadruple the convenient number 60 to get 240. Adding 240 to 3,600 results in 3,840. Finally we add 4 to get 3,844 = 382.

Let's calculate 472. Two less than the number is 45, and 452 = 2,025 by the rule of squaring a number ending in 5. We then quadruple 45, which is 180. Adding 180 to 2,025 results in 2,205, and then adding 4 means 472 = 2,209.

In the cases of 8 or 3, as well as 7 or 2, you always square the convenient number and then add 4 at the end. If the convenient number is larger than the original number (8 or 3), then you need to subtract quadruple the convenient number. If the convenient number is smaller than the original number (7 or 2), then you need to add quadruple the convenient number.

Practice Problems

182

432

772

1222

2522

Proof

Here is a proof for the rule of 8 or 3. For any number x, you can obtain x2 by squaring x + 2, subtracting 4 times x + 2, and then adding 4. The algebraic proof is (x + 2)2 – 4(x + 2) + 4 = (x2 + 4x + 4) – 4x – 8 + 4 = x2.

Here is a proof for the rule of 7 or 2. For any number x, you can obtain x2 by squaring x – 2, adding 4 times x – 2, and then adding 4. The algebraic proof is (x – 2)2 + 4(x – 2) + 4 = (x2 – 4x + 4) + 4x – 8 + 4 = x2.

Solutions to Practice Problems

182. We first calculate 202 = 400. Then we subtract 4 x 20 = 80 and add 4, so we have 182 = 400 – 80 + 4 = 324.

432. We calculate 452 first by the rule of squaring a number ending in 5. This is 452 = 2,025. Then we subtract 4 x 45 = 2 x 90 = 180 and then add 4. So we have 432 = 2,025 – 180 + 4 = 1,849.

772. We need to calculate 752 first. This is found by the rule of squaring a number ending in 5. So we calculate 7 x 8 = 56 and then append the digits 25, so 752 = 5,625. Then we need to add 4 x 75 = 2 x 150 = 300 and add 4. So we have 772 = 752 + 300 + 4 = 5,625 + 304 = 5,929.

1222. We calculate 1202 = 14,400 first. Then we add 4 x 120 = 480 and 4 to get 1222 = 14,400 + 480 + 4 = 14,884.

2522. We need to calculate 2502 first which is 62,500 (this can be found because 2502 = (25 x 10)2 = 252 x 102 = 252 x 100, and 252 can be found by the method for squaring a number ending in 5). Then we need to add 4 x 250 = 1,000 and 4. Thus 2522 = 2502 + 1,000 + 4, and finally this gets to the answer of 62,500 + 1,004 = 63,504.



Square A Two-Digit Number General Procedure

There is an alternate method to squaring any two-digit number if you are comfortable doing a few calculations in your head involving carrying over. Let's do an example and then explain the procedure.

What is the square of 23? We add the second digit to the original number. So we add 3 to 23 to get 26. Then we multiply this by the first digit of 2 to get 52, which we append a 0 to get 520. We finally add in the square of the second digit 3, which is 9. The result is 529.

The procedure breaks down squaring the number into the following easier calculations: add the second digit to the original number, multiply by the first digit, multiply by 10, and then finally add the square of the second digit.

Let's try another example of the square of 72. We add 2 to 72 to get 74, which is then multiplied by 7 to get 518. We multiply by 10 to get 5,180. Finally we add the square of 2, which is 4, to get 5,184.

Practice Problems

282

342

172

862

912

Proof

A two-digit number x can be expressed as x = 10a + b. From algebra, we have x2 = (10a + b)2 = 100a + 20ab + b2 = 10a(10a + 2b) + b2, which equals 10a(x + b) + b2. The term x + b means to add the second digit of b to the number x. That is multiplied by the first digit of a, which is further multiplied by 10. Finally the square of second digit b is added.

Solutions to Practice Problems

282. We add 8 to 28 to get 36, then that is multiplied by 2 to get 72, and that is further multiplied by 10 to get 720. Then we add 82 = 64 to get the answer 784.

342. We add 4 to 34 to get 38, then that is multiplied by 3 to get 114, and that is further multiplied by 10 to get 1,140. Then we add 42 = 16 to get the answer 1,156.

172. We add 7 to 17 to get 24. Multiplying by 1 is still 24, and that is further multiplied by 10 to get 240. Then we add 72 = 49 to get the answer 289.

862. We add 6 to 86 to get 92. Now we need to multiply 92 by 8. We can instead do 92 times (10 – 2) to get 920 – 184 = 736. That is multiplied by 10 to get 7,360. Then we add 62 = 36 to get the answer 7,396.

912. We add 1 to 91 to get 92, then that is multiplied by 9 to get 828, and further multiplied by 10 to get 8,280. Then we add 12 = 1 to get the answer 8,281.



Square A Number Between 41 And 59

How would you square 54? Many methods have been explained already. There is in fact a very quick way to do this problem that is worth learning.

The first step is to consider the difference between 54 and 50. The difference is +4. The last two digits of our answer are the square of the difference. Since 4 squared is equal to 16, the last two digits of our answer are 16.

The first two digits of the answer will be half of 50, which is 25, plus the difference of +4. Since 25 + 4 = 29, the first two digits of the answer are 29.

Putting this together, the answer is 2,916.

We can similarly square any number from 51 to 59 in the following process. The first two digits are equal to 25 plus the difference, and the last two digits are equal to the difference squared. For example, 58 has a difference of +8. The first two digits will be 25 + 8 = 33 and the last two digits will be 82 = 64. So the answer is 3,364.

How would we find the square of 47? The process is almost the same. The only change that the number 47 is less than 50. Therefore, we need to subtract the difference from 25.

Here's how we can square 47. We notice the number 47 is 3 less than 50. The first two digits of our answer are 25 – 3 = 22. The last two digits are the square of 3, which we write as a two-digit number 09. Thus we have 2,209 as the answer.

You can also apply this rule to remember the square of 50. This number has a difference of 0 from 50, and so the first two digits are 25 + 0 = 25, and the last two digits are the square of 0, which is 00 when written as a two-digit number. Thus the answer is 2,500.

To summarize, you can square a number from 41 to 59 as follows. Adjust the number 25 by the difference from 50 to get the first two digits of the answer. Then, square the difference, and write that as a two-digit number, to get the last two digits of the answer.

Practice Problems

442

572

462

432

522

Proof

For any number x, (50 + x)2 = 2,500 + 100x + x2 = 100(25 + x) + x2. The term 25 + x means to adjust the number 25 by x, which is the difference from 50. This is multiplied by 100 to become the first two digits of the answer. Then the square of the difference is added, and those two digits are the last two digits of the answer.

Solutions to Practice Problems

442. The number 44 is 6 less than 50. So we subtract 6 from 25 to get 19, and then append 62 = 36. The result is 1,936.

572. The number 57 is 7 more than 50. So we add 7 to 25 to get 32, and then append 72 = 49. The result is 3,249.

462. The number 46 is 4 less than 50. So we subtract 4 from 25 to get 21, and then append 42 = 16. The result is 2,116.

432. The number 43 is 7 less than 50. So we subtract 7 from 25 to get 18, and then append 72 = 49. The result is 1,849.

522. The number 52 is 2 more than 50. So we add 2 to 25 to get 27, and then append 22 = 04 (remembering to write the square as a two-digit number!). The result is 2,704.



Square A Number Around 500

How would you square 504? There is a general rule similar to the one we just explained for squaring a number around 50.

The first step is to consider the difference between 504 and 500. The difference is +4. The last three digits of our answer—since 500 is a three-digit number—will be the square of the difference. Since 4 squared is equal to 16, the last three digits of our answer are 016.

The first three digits of the answer will half of 500, which is 250, plus the difference of 4. Since 250 + 4 = 254, the first three digits of the answer are 254. Putting this together, the answer is 254,016.

We can similarly square any number above 500 by the same process. The first three digits of the answer will be 250 plus the difference, and the last three digits will be the difference squared, written as a three-digit number. For example, 5062 can be calculated as 250 + 6 = 256 (the first three digits) and 62 = 36 = 036 (the last three digits). So the answer is 256,036.

We can similarly find the squares of numbers less than 500. We just need to remember to subtract the difference from 250 since these numbers are less than 500. So let's square 498. The difference of this number is 2. The first three digits will be 250 – 2 = 248, and the last three digits will be 2 squared, which is 004 when written as a three-digit number. So 498 squared is equal to 248,004.

You can generalize this process to find the square near any number that starts with a 5 and ends with 0s. For example, let's do 5,000. Since 5,000 is a four-digit number, the first four digits will be 2,500 plus/minus the difference, and the final four digits will be the difference squared. To square numbers near 50,000, it's the same rule except we now have a five digit number. Therefore the first five digits will be 25,000 plus/minus the difference, and the final five digits will be the difference squared.

You can see how the pattern continues. The number with 25 is always half of the original number. So with 50 the number is 25, then with 500 it is 250, with 5,000 it is 2,500, and so on.

Practice Problems

5052

5072

4962

4932

5,0022

Proof

A number near 50, 500, 5,000, or so on, can be written as 5(10)k + x. To square the number, we have (5(10)k + x)2 = 25(10)2k + (10)k+1x + x2. This equals (10)k+1(25(10)k-1 + x) + x2. In other words, we add the difference x to the number 25 with appropriate number of zeros to become the leading digits of the answer. Then we add the x2 term, written with the appropriate number of digits, to become the ending digits of the answer.

Solutions to Practice Problems

5052. The number 505 is 5 more than 500. So we add 5 to 250 to get 255, and then append 52 = 025 (remembering to write as a three-digit number). The result is 255,025.

5072. The number 507 is 7 more than 500. So we add 7 to 250 to get 257, and then append 72 = 049. The result is 257,049.

4962. The number 496 is 4 less than 500. So we subtract 4 from 250 to get 246, and then append 42 = 016. The result is 246,016.

4932. The number 493 is 7 less than 500. So we subtract 7 from 250 to get 243, and then append 72 = 049. The result is 243,049.

5,0022. The number 5002 is 2 more than 5,000. So we add 2 to 2,500 to get 2,502, and then append 22 = 0004 (with four-digit numbers we need this square to be written as a four-digit number). The result is 25,020,004.



Square A Two-Digit Number Faster

There is a quicker way to square any two-digit number, if you first memorize the squares up to 25.

12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25

62 = 36, 72 = 49, 82 = 64, 92 = 81, 102 = 100

112 = 121, 122 = 144, 132 = 169, 142 = 196, 152 = 225

162 = 256, 172 = 289, 182 = 324, 192 = 361, 202 = 400

212 = 441, 222 = 484, 232 = 529, 242 = 576, 252 = 625

The squares from 26 to 99 can be calculated from these memorized values.

For the numbers between 26 and 75, the method is to use the rule for squaring numbers around 50. We find the difference from 50, and we either subtract that (for numbers less than 50) or we add that (for numbers larger than 50) from the number 25. These are the first two digits of the answer. Then we square the difference and add that as the last two digits. If the square of the difference is three digits, then we have to carry over to the first two digits.

Let's do an example of 382. This number is 12 less than 50, so we subtract 12 from 25 to get 13. These are the first two digits of the answer. Then we square the difference of 12 to get 144. The 44 are the last two digits, and the 1 carries over to the 13 to become 14. Therefore the answer is 1,444.

Let's do another example of 732. This number is 23 more than 50, so we add 23 to 25 to get 48. These are the first two digits of the answer. Then we square the difference of 23 to get 529 (this is why the squares up to 25 have to be memorized). The 29 are the last two digits, and the 5 carries over to the 48 to become 53. Therefore the answer is 5,329.

For numbers 76 to 99, we can look at the difference from 100. The procedure is to subtract double the difference from 100, and then add the square of the difference, with carry-over if the square is three digits or more.

For example, let's do 872. The difference from 100 is 13, so we subtract double the difference of 26 from 100 to get 74. These are the first two digits. Then we square the difference of 13 to get 169. Since this is a three-digit result, we carry over the 1 to the 74 to make 75. So the answer is 7,569.

For another example, let's do 792. The difference from 100 is 21, so we subtract double the difference of 42 from 100 to get 58. These are the first two digits. Then we square the difference of 21 to get 441. Since this is a three-digit result, we carry over the 4 to the 58 to make 62. So the answer is 6,241.

In fact, you can quickly adapt this trick to do squares between 101 and 125. The rule is to add double the difference to 100, and then add the square of the difference.

For instance, 1122 can be calculated as 100 + 2(12) = 124 as the first three digits, and then 122 = 144. We carry over the 1 to make 124 into 125. So the answer is 12,544.

Practice Problems

362

292

672

912

1172

Proof

The rule for numbers 26 to 75 works for the same reason as squaring a number between 41 and 59, which was already proven.

For numbers between 76 and 100, the rule works because for any number x, we have (100 – x)2 = 10,000 – 200x + x2 = 100(100 – 2x) + x2. That is, we subtract the double difference x from 100, shift over by 2 decimal places, and then add x2.

For numbers 101 to 125, the rule is to add double the difference to 100, and then add the square of the difference. This works because for any number x, we have (100 + x)2 = 10,000 + 200x + x2 = 100(100 + 2x) + x2.

Solutions to Practice Problems

362. The number 36 is 14 less than 50. So we subtract that from 25 to get 11. Then we add 142 = 196. The digits 96 become the last two digits, and the 1 carries over to the 11 to make 12. So the answer is 1,296.

292. The number 29 is 21 less than 50. So we subtract that from 25 to get 4. Then we add 212 = 441. The digits 41 become the last two digits, and the hundreds digit 4 carries over to the 4 to make 8. So the answer is 841.

672. The number 67 is 17 more than 50. So we add that to 25 to get 42. Then we add 172 = 289. The digits 89 become the last two digits, and the hundreds digit 2 carries over to the 42 to make 44. So the answer is 4,489.

912. The number 91 is 9 less than 100. So we subtract double the amount, 18, from 100 to get 82. Then we add 92 = 81. The result is 8,281.

1172. The number 117 is 17 more than 100. So we add double the amount, 34, to 100 to get 134. Then we add 172 = 289. The hundreds digit of 2 carries over to 134 to make 136. The digits 89 are the last two digits of the answer. The result is 13,689.



Square 34, 334, Etc. Or Square 67, 667, Etc.

This is a fairly useless math trick, but there is a mathematical reason for why this trick was discovered, which is why this trick is mentioned.

There is a pattern to squaring numbers starting with a series of 3s and ending in a 4. Here are some examples.

342 = 1,156

3342 = 111,556

3,3342 = 11,115,556

For a number starting with N digits of a 3 and ending in a 4, the square of that number will have N + 1 digits of 1, then N digits of 3, and the last digit will be a 6.

Similarly, there is a pattern to squaring numbers starting with a series of 6s and ending in a 7. Here are some examples.

672 = 4,489

6672 = 444,889

6,6672 = 44,448,889

For a number starting with N digits of a 6 and ending in a 7, the square of that number will have N + 1 digits of 4, then N digits of 8, and the last digit will be a 9.

So why are these numbers important? In 1985, Donald E. Knuth held one of his regular problem-solving sessions at Stanford University. He asked if there were any numbers x and x2 where the digits in both numbers were non-decreasing. Mathematicians sometimes ask these kinds of questions out of curiosity or to identify patterns. Occasionally, there are surprising results that have practical value, or the proof might involve a method of reasoning that is useful for solving other problems. So while you might find this question a bit strange, it is the kind of exercise that helps mathematically minded individuals sharpen their skills.

It was discovered numbers of the form 33...4 and 66...7 have the property, as it is easy to see the digits in the numbers are non-decreasing and the digits in the squares are also non-decreasing. Since there are infinitely many numbers of these forms, they could conclude there were infinitely many numbers for which x and x2 have non-decreasing digits.

For more details, you can read the transcript of the 1985 session with Donald Knuth (search for “monotonic squares” or go to page 9) online: http://www-cs-faculty.stanford.edu/~uno/papers/cs1055.pdf

Practice Problems

33,3342

66,6672

Proof

Here is a proof for the pattern of 33...4 squared. A number starting with 3s and ending in a 4 is equal to 10n+1/3 + 2/3. Thus,

(33...4)2 = (10n+1/3 + 2/3)2 = 102n+2/9 + 10n+1(4/9) + 4/9

When we divide a power of 10 by 9, we get a number with only 1s plus a fraction of 1/9. So the first term has the digit 1 repeated 2n + 2 times plus 1/9, and the middle term has the digit 4 repeated n + 1 times plus 4/9.

(33...4)2 = (1...1 + 1/9) + (4...4 + 4/9) + 4/9

Now we add up the terms. There are n + 1 leading digits of 1; the next n digits are 5 = 1 + 4; and the final digit is equal to 5 plus the fractions 1/9, 4/9, and 4/9, which neatly sum to 1. Thus the final digit is 6. So in the end, we have the result as desired.

(33...4)2 = 1...14...45

Where there are n + 1 leading digits of 1, then n digits of 4, and a final digit of 5.

The proof for 66...7 is similar, using the fact that a number 66...7 can be written as 10n+1(2/3) + 1/3. So when we square the number, we have the following equation.

(66...7)2 = (10n+1(2/3) + 1/3)2 = 102n+2(4/9) + 10n+1(4/9) + 1/9

When we divide a power of 10 by 9, we get a number with only 1s plus a fraction of 1/9. So the first term has the digit 4 repeated 2n + 2 times plus 4/9, and the middle term has the digit 4 repeated n + 1 times plus 4/9.

(66...7)2 = (4...4 + 4/9) + (4...4 + 4/9) + 1/9

Now we add up the terms. There are n + 1 leading digits of 4; the next n digits are 8 = 4 + 4; and the final digit is equal to 8 plus the fractions 4/9, 4/9, and 1/9, which neatly sum to 1. Thus the final digit is 9. So in the end, we have the result as desired.

(33...4)2 = 4...88...89

Where there are n + 1 leading digits of 4, then n digits of 8, and a final digit of 9.

Solutions to Practice Problems

33,3342. We need to write one more digit of 1 as there are digits of 3s, so the result starts out with five 1s. Then we write down as many 5s as there are 3s, so there are four 5s. And finally there is a 6. So the answer is 1,111,155,556.

66,6672. We need to write one more digit of 4 as there are digits of 6s, so the result starts out with five 4s. Then we write down as many 8s as there are 6s, so there are four 6s. And finally there is a 9. So the answer is 4,444,488,889.





Part III: Multiplying Numbers


Multiply Two Numbers Ending In 5

What is 45 x 65? There is a shortcut to multiplication problems when both numbers end in the digit 5.

There are four steps. First, multiply the tens digits. Second, add the average of the tens digits. Third, multiply the result by 100. And finally add 25.

For 45 x 65, we first multiply the tens digits of 4 and 6 to get 24. Then we add the average of 5 to get 29. Multiplying by 100 results in 2,900, and finally adding 25 gets the result of 2,925.

Let's do another example of 35 x 75. We first multiply the tens digits of 3 and 7 to get 21. Then we add the average of 5 to get 26. Multiplying by 100 results in 2,600, and finally adding 25 gets the result of 2,625.

Let's do a final example of 55 x 85, we first multiply the tens digits of 5 and 8 to get 40. Then we add the average, which is (5 + 8)/2 = 6.5. It is okay that the average is not a whole number. Adding this to the product gets 46.5. Then we multiply by 100 to get 4,650, and finally adding 25 gets the result of 4,675.

The rule can also be extended to three-digit numbers or more. In that case, the rule is the same, with the “tens” digit being the number obtained by deleting the 5. For example, in the number 125, the “tens” digit would be the number 12.

The rule is sometimes written differently. If the numbers are a5 and b5, you can obtain the leading digits by adding the product ab to the average, (a + b)/2, rounded down. If a and b are both even or both odd, then the last two digits are 25. If one is one and one is even, the last two digits will be 75.

For example, let's do 55 x 85. The first part is the product 5 x 8 = 40 plus the average (5 + 8)/2 = 6.5, rounded down to 6. So we add 40 and 6 to get 46. Since 5 is odd and 8 is even, the last two digits will be 75. So the answer is 55 x 85 = 4,675. This is exactly the same result as before, but some people find this procedure easier to learn or quicker to use.

Practice Problems

15 x 35

85 x 65

35 x 45

95 x 5

105 x 145

Proof

Two numbers ending in 5 are written as (10a + 5) and (10b +5). When we multiply them, we have (10a + 5)(10b + 5) = 100ab + 50(a + b) + 25, which is equal to 100[ab + (a + b)/2] + 25. That is, we compute the product plus the average as ab + (a + b)/2, multiply by 100, and then add 25.

Solutions to Practice Problems

15 x 35. The tens digits are 1 and 3. We add the product 1 x 3 = 3 and the average, 2, to get 5. Then we multiply by 100 to get 500 and add 25 to get 525.

85 x 65. The tens digits are 8 and 6. We add the product 8 x 6 = 48 and the average, 7, to get 55. Then we multiply by 100 to get 5,500 and add 25 to get 5,525.

35 x 45. The tens digits are 3 and 4. We add product of 3 x 4 = 12 and the average, 3.5, to get 15.5. Then we multiply by 100 to get 1,550 and add 25 to get 1,575.

95 x 5. We can write this problem as 95 x 05. The tens digits are 9 and 0. We add the product 9 x 0 = 0 and the average, 4.5, to get 4.5. Then we multiply by 100 to get 450 and add 25 to get 475.

105 x 145. The “tens” digits are found by deleting the 5 in each number to get the numbers 10 and 14. We add the product 10 x 14 = 140 and the average, 12, to get 152. Then we multiply by 100 to get 15,200 and add 25 to get 15,225.



Multiply Numbers Differing By 2

My friend was asked the following question in an investment bank interview: what is 14 times 16? The interviewer then put pressure on her to answer quickly. How was she supposed to do this without a calculator?

The trick was to see 14 x 16 is the same as (15 – 1)(15 + 1), and the product of (x + y)(x – y) is equal to x2 – y2. In other words, she was supposed to see that 14 x 16 is also equal to 152 – 12 = 225 – 1 = 224.

We can use this trick for any numbers that differ by 2. The product of the two numbers will be the square of their average minus 1.

To do another example, 7 x 9 is equal to 82 –1. And in fact we can verify 7 x 9 = 63 = 82 – 1 = 64 – 1.

We can also combine this trick with the methods on how to square numbers quickly. For example, what is 51 x 53? Because 51 and 53 differ by 2, we know their product is equal to the square of their average of 52 minus 1. That is, 51 x 53 is equal to 522 – 1. But what is 52 squared? We can use the trick of squaring numbers between 41 and 59. We add the difference of 2 to 25 to get 27, then we append the square of 2 written as a two-digit number, 04. So the square of 52 is 2,704, and therefore 51 x 53 = 2,704 – 1 = 2,703.

This rule can be extended to numbers that differ by 4, 6, or any even number. That is, the rule works whenever two numbers are equidistant from the same number. Their product will be the square of their average minus the square of their distance from the average. Note the distance to the average is half of the difference between the two numbers.

For example, what is 14 x 18? These numbers are equidistant from 16, with a distance of 2 from the average. Therefore, their product is equal to the square of 16 minus the square of 2. That is, 14 x 18 = 162 – 22, which is equal to 256 – 4 = 252.

Similarly, we can calculate 13 x 21. The average of the two numbers is 17, and they are a distance of 4 from the average. Therefore, their product is equal to the square of 17 minus the square of 4. That is, we can compute 13 x 21 = 172 – 42, which is equal to 289 – 16 = 273.

Practice Problems

11 x 13

18 x 16

13 x 19

98 x 102

46 x 54

Proof

Numbers at distance y from x are (x – y) and (x + y). The product of these numbers is (x – y)(x + y) = x2 – y2. That is, when two numbers are a distance y from their average x, their product is the square of the average minus the square of the distance from the average.

Solutions to Practice Problems

11 x 13. We re-write 11 x 13 = (12 – 1)(12 + 1), and then solve 122 – 12, which is equal to 144 – 1 = 143.

18 x 16. We re-write 18 x 16 = (17 + 1)(17 – 1), and then solve 172 – 12, which is equal to 289 – 1 = 288.

13 x 19. We re-write 13 x 19 = (16 – 3)(16 + 3), and then solve 162 – 32, which is equal to 256 – 9 = 247.

98 x 102. We re-write 98 x 102 = (100 – 2)(100 + 2). Then we can solve 1002 – 22 = 10,000 – 4 = 9,996.

46 x 54. We re-write 46 x 54 = (50 – 4)(50 + 4), and then solve 502 – 42, which is equal to 2,500 – 16 = 2,484.



Multiply Numbers Units Digits Sum To 10 And Same Tens Digit

In the last section, we solved 14 x 16 = 152 – 12. This is a good way to solve the problem. But there is an alternate way that is perhaps even faster! This is one of my favorite tricks because the problems can be solved very easily.

You do, however, have to check the conditions for this trick. What you need is the units digits to sum to 10 and the tens digits to be the same. For example, in 14 x 16, the units digits of 4 and 6 do sum to 10, and the tens digit of 1 is the same for both numbers. This trick will when the units digits come from the pairs (1, 9), (2, 8), (3, 7), or (4, 6). It will also work for the pair (5, 5). In that case, both numbers will be the same, and the trick will be equivalent to squaring a number ending in 5.

While the trick has several conditions, these problems do arise, and there is a really fast and fun way to solve them.

Here is the procedure. You first multiply the tens digit by one more than itself. Then you multiply the units digits together and append the result, written as a two-digit number. And that's your answer!

For example, let's do 14 x 16. We multiply the tens digit of 1 by one more than itself, 2. So we have 1 x 2 = 2. Then we multiply the units digits of 4 and 6. This is 24, which we append to 2. So our answer is 224. That was easy!

Let's do another example. Let's try 33 x 37. We multiply the tens digit of 3 by one more than itself, 4. This is 3 x 4 = 12. Then we multiply the units digits of 3 and 7 to get 21, which we append to 12. Putting it together, we have 33 x 37 = 1,221. This is pretty cool!

You can also use the trick for three-digit or larger numbers. In that case, you need all the digits except the units digit to be the same (you can consider that the “tens” digit part). Then the procedure is the same: you multiply the “tens” digit part by one more than itself, and then you append the product of the units digit as the last two digits of the answer.

Let's do an example of 121 x 129. The “tens” digit in both numbers is the part not in the units digit. In both numbers the “tens” digit is 12. So to do the trick, we first multiply 12 by one more than itself, 13, to get 156. Now we need to multiply the units digits of 1 and 9, which we write as a two-digit number 09. Putting it together, we have 121 x 129 = 15,609.

The trick is very easy to do and impressive when you can pull it off, so keep an eye out for these types of problems.

(Incidentally, if you did not know how to do 12 x 13, there are tricks we have learned to solve this problem. One method is the trick to multiply numbers between 11 and 19. We add the units digit of 3 to the number 12 to get 15. This is multiplied by 10 to get 150. And then we would add the product of the units digits 2 x 3 = 6 to get the answer of 12 x 13 = 156. Another method is to split up the multiplication into parts by seeing that 13 = 12 + 1. Then 12 x 13 = 12 x (12 + 1) = 122 + 12 = 144 + 12 = 156.)

Practice Problems

12 x 18

21 x 29

43 x 47

98 x 92

56 x 54

Proof

Write one number as 10a + b. The units digit of the other number should be 10 – b so that the units digits sum to 10. The tens digit should be the same, so the tens digit is 10a. So the other number is 10a + (10 – b).

Multiplying the numbers 10a + b and 10a + (10 – b), we get the result of (10a + b)[10a + (10 – b)] = 100a2 + 10a(10 – b) + b(10a) + b(10 – b). We can cancel terms and group the remaining terms. Ultimately, this expression simplifies to 100a(a + 1) + b(10 – b). The term a(a + 1) means to multiply the leading digit a by one more than itself, and that is multiplied by 100 so it starts in the hundreds spot. The term b(10 – b) means to multiply the two units digits together, and that result becomes the last two digits of the answer.

Solutions to Practice Problems

12 x 18. We multiply 1 by one more than itself, 2, to get 2. Then we multiply the units digits, 2 x 8 = 16, to get the last two digits. The answer is 216.

21 x 29. We multiply 2 by one more than itself, 3, to get 6. Then we multiply the units digits, 1 x 9 = 9, to get the last two digits of 09. The answer is 609.

43 x 47. We multiply 4 by one more than itself, 5, to get 20. Then we multiply the units digits, 3 x 7 = 21, to get the last two digits. The answer is 2,021.

98 x 92. We multiply 9 by one more than itself, 10, to get 90. Then we multiply the units digits, so 8 x 2 = 16, to get the last two digits. The answer is 9,016.

56 x 54. We multiply 5 by one more than itself, 6, to get 30. Then we multiply the units digits, so 6 x 4 = 24, to get the last two digits. The answer is 3,024.



Multiply Numbers Tens Digits Sum To 10 And Same Units Digit

What is 34 x 74? There is a complement trick when the tens digits sum to 10 and the units digits are the same. For example, in 34 x 74, the units digits are both 4, while the tens digits of 3 and 7 sum to 10. This trick applies if the tens digits belong to the pairs (1, 9), (2, 8), (3, 7), or (4, 6). It will also work for the pair (5, 5), but in that case you will be squaring a number in the 50s, and it would be quicker to use that trick.

When the two numbers meet these conditions, you can solve the problem very quickly.

Here is the procedure. You multiply the tens digits together and then add the units digit. In 34 x 74, we calculate 3 x 7 = 21, and then we add 4 to get 25. These are the first two digits. Then we multiply the units digits (equivalent to squaring the units digit), and those are the last two digits. Returning to the example, 4 x 4 = 42 = 16. Appending 16 to 25 gets the answer 2,516.

Let's do another example of 46 x 66. We multiply the tens digits of 4 and 6 to get 24, to which the units digit of 6 is added to get 30. Then we square 6 to get 36. Appending 36 to 24 gets the answer 3,036.

Practice Problems

21 x 81

12 x 92

34 x 74

89 x 29

65 x 45

Proof

Write one number as 10a + b. The units digit of the other number must be the same number b. The tens digit of the other number must sum to 10, so the tens digit should be 10 – a. This means the other number is 10(10 – a) + b.

Multiplying the numbers of 10 – a and 10(10 – a) + b, we get the result of (10a + b)[10(10 – a) + b] = 100a(10 – a) + 10ab + 10b(10 – a) + b2. We can cancel terms and group them to get 100[a(10 – a) + b] + b2. The term a(10 – a) + b means to multiply the tens digits together and then add the units digit. That is multiplied by 100, which shifts the value over to the hundreds spot. The term b2 means to multiply the units digits together, and those are the last two digits of the answer.

Solutions to Practice Problems

21 x 81. Multiply the tens digits together, 2 x 8 = 16, and then add the units digit of 1 to get 17 as the first two digits. Then square the units digit, 12 = 1, and write that as the two-digit number 01. Appending 01 to 17 gets the result 1,701.

12 x 92. Multiply the tens digits together, 1 x 9 = 9, and then add the units digit of 2 to get 11 as the first two digits. Then square the units digit, 22 = 4, and write that as the two-digit number 04. Appending 04 to 11 gets the result 1,104.

34 x 74. Multiply the tens digits together, 3 x 7 = 21, and then add the units digit of 4 to get 25 as the first two digits. Then square the units digit, 42 = 16. Appending 16 to 25 gets the result 2,516.

89 x 29. Multiply the tens digits together, 8 x 2 = 16, and then add the units digit of 9 to get 25 as the first two digits. Then square the units digit, 92 = 81, and write that as the two-digit number 81. Appending 85 to 25 gets the result 2,581.

65 x 45. Multiply the tens digits together, 6 x 4 = 24, and then add the units digit of 5 to get 29 as the first two digits. Then square the units digit, 52 = 25. Appending 25 to 29 gets the result 2,925.



Multiply A Two-Digit Number By 11

What is 34 multiplied by 11? Here's a trick to solve the problem quickly.

There are three steps. First, copy the leading digit 3. Then, add up the two digits, 3 + 4 = 7. Finally, copy the second digit 4. Those numbers, written in order, are the digits of the answer. So 34 x 11 = 374.

To summarize, we can multiply a two-digit number by 11 by copying the first digit, adding the two digits together, and then copying the second digit. Those numbers written in order are the digits of the answer.

What about 57 times 11? We copy the first digit of 5. Then we have to add 5 and 7 which is 12. When the sum of the two digits is 10 or larger, we have to carry over the tens part to the previous digit of the answer. That is, from the sum of 12, we carry over the tens digit of 1 to make the previous digit of 5 into a 6. So we have 62 as the first two digits of the answer. Finally we copy the last digit of 7 to get 627.

Let's do another example. Let's multiply 76 times 11. We copy the first digit of 7. When we add 7 and 6 to get 13. Carrying over the tens digit of 1, the previous digit of 7 becomes an 8. So we have 83 as the first two digits of the answer. Finally we copy 6 as the last digit to get 836.

You can either carry over as you do the middle sum, or you can make the adjustment at the end. For example, we could do 38 x 11 by noting the answer is 3, 3 + 8, 8 = 3, 11, 8 and then carry over the 11 term to get the digits 4, 1, 8. Then we know 418 is the answer.

Practice Problems

21 x 11

45 x 11

81 x 11

56 x 11

97 x 11

Proof

A two-digit number can be written as 10a + b. Multiplying by 11 is the same as multiplying by (10 + 1).

This means (10a + b)11 = (10a + b)(10 + 1) = 100a + 10b + 10a + b, which is equal to 100a + 10(a + b) + b. So we copy the digit a to the hundreds spot, then we take the sum a + b as the tens spot, and finally we copy the digit b to the units spot.

Solutions to Practice Problems

21 x 11. We copy the digit 2, then we add 2 + 1 = 3, and then we copy the last digit 1. So the answer is 231.

45 x 11. We copy the digit 4, then we add 4 + 5 = 9, and then we copy the last digit 5. So the answer is 495.

81 x 11. We copy the digit 8, then we add 8 + 1 = 9, and then we copy the last digit 1. So the answer is 891.

56 x 11. We copy the digit 5. Then we add 5 + 6 = 11. The next digit is 1 and we carry over the 1 so the 5 becomes a 6. So our answer starts out as 61. Then we copy the last digit 6. So the result is 616.

97 x 11. We copy the digit 9. Then we add 9 + 7 = 16. The next digit is 6 and we carry over the 1 so the 9 becomes a 10. So our answer starts out as 106. Then we copy the last digit 7. So the result is 1,067.



Multiply Any Number By 11

What is 1,234 multiplied by 11? We can use a procedure similar to multiplying a two-digit number by 11.

The general process is: copy the first digit, add each pair of consecutive digits, and copy the last digit. If any sum is 10 or more, then carry over the digit in the tens spot to the previous digit of the answer.

Let's use this rule to multiply 1,234 by 11. We copy the first digit 1. Then we add each pair of consecutive digits. So we have 1 + 2 = 3, then we go to the next pair 2 + 3 = 5, and then 3 + 4 = 7. This means our middle digits are 357. Finally we copy the last digit of 4. Putting this all together, the answer is 13,574.

Let's do 567 times 11. We copy the first digit 5. Then we add each pair of consecutive digits. So we have 5 + 6 = 11, and 6 + 7 = 13. Then we copy the last digit of 7. So the answer has the digits 5, 11, 13, 7. Now we carry over for any digits that are 10 or larger. We carry over the 1 from the 11 so the digits are 6, 1, 13, 7. Then we carry over the 1 from the 13 so the digits are 6, 2, 3, 7. Now every digit is 9 or smaller, so these are the digits of the answer. Thus, 567 x 11 = 6,237.

As this example illustrates, it is a bit harder to use this method when the digits get larger and you need to carry over many times. It can be helpful to do the sums of consecutive numbers and then account for the carryover at the end.

Practice Problems

121 x 11

245 x 11

381 x 11

12,597 x 11

8,456 x 11

Proof

A number with k digits is written as 10ka + 10k-1b + 10k-2c + … + z. Multiplying by 11 is the same as multiplying by 10 and then adding 1. Multiplying by 10 increases the power of 10 in each term. So the expression is 10k+1a + 10kb + 10k-1c + … + 10z. Then we add back the original number. If we group the terms by the powers of 10, we end up with 10k+1a + 10k(a + b) + 10k-1(b + c) + … + 10(y + z) + z. In other words, the highest power of 10 is copied from the first digit of the number a. Then each term is the sum of consecutive digits. And finally the last digit z is copied as the last digit of the answer.

Solutions to Practice Problems

121 x 11. We copy the digit 1. Then 1 + 2 = 3, and 2 + 1 = 3. Finally we copy 1 as the last digit. So the answer is 1,331.

245 x 11. We copy the digit 2. Then 2 + 4 = 6, and 4 + 5 = 9. Finally we copy 5 as the last digit. So the answer is 2,695.

381 x 11. We copy the digit 3. Then 3 + 8 = 11. We need to carry over the 1 so the previous digit 3 becomes a 4. So the answer starts out as 41. Then 8 + 1 = 9, and finally we copy 1 as the last digit. So the answer is 4,191.

12,597 x 11. This example also requires many carry-overs. So we'll do it without the carry-over and then make adjustments. We first copy 1, then we do the sums of 1 + 2 = 3, 2 + 5 = 7, 5 + 9 = 14, 9 + 7 = 16, and copy the last digit 7. So the results are 1, 3, 7, 14, 16, 7. We first carry over the 1 from the 14 to make 1, 3, 8, 4, 16, 7. Now we carry over the 1 from the 16 to make 1, 3, 8, 5, 6, 7. Now every digit is 9 or less, so the answer is 138,567.

8,456 x 11. This example requires many carry-overs. So we'll do it without the carry-over and then make adjustments. We first copy 8, then we do the sums of 8 + 4 = 12, 4 + 5 = 9, and 5 + 6 = 11, and copy the last digit 6. So the results are 8, 12, 9, 11, 6. We first carry over the 1 from the 12 to make 9, 2, 9, 11, 6. Now we carry over the 1 from the 11, so this increases the 9 into a 10, and we have 9, 2, 10, 1, 6. Now we carry over the 1 from the 10, so the 2 increases to a 3. This means 9, 3, 0, 1, 6. Now every digit is 9 or less, so the answer is 93,016.



Multiply Two Numbers In The 90s

Can you multiply 95 and 96 in your head? There is a wonderfully easy way to do this and similar problems.

There are three steps to the method. First, consider the difference of each number from 100. Second, subtract the difference of one number from the other number to get the first two digits of the answer. Finally, multiply the differences to get the last two digits of the answer.

For example, let's do 95 x 96. The first step is to consider the differences of each number from 100. The number 95 is 5 less than 100, and the number 96 is 4 less than 100. We then subtract the difference of one number from the other number. So we subtract 4 from 95 to get 91. We could also have subtracted 5 from 96 to get 91. Notice that both are the same so it doesn't matter which way we do it. This result of 91 becomes the first two digits of the answer. Finally, we multiply the differences. So we multiply 5 by 4 to get 20. These are the last two digits of the answer. Putting it together, we have 95 times 96 is equal to 9,120.

Let's do another example of 97 times 98. The first step is to consider the differences from 100. The number 97 is 3 less than 100 and the number 98 is 2 less than 100. Second, we subtract the difference of one number from the other number. So we have 95 = 97 – 2 = 98 – 3. The final two digits can be found by multiplying the differences together. So we have 2 times 3 which equals 6. We have to remember to write this as a two-digit number, meaning the final two digits are 06. Putting the steps together, we have 97 times 98 is equal to 9,506.

Practice Problems

95 x 98

96 x 97

98 x 99

91 x 92

942

Proof

Numbers in the 90s can be written 100 – x and 100 – y for x and y less than 10. The product is (100 – x)(100 – y) = 10,000 – 100y – 100x + xy, which is equal to 100(100 – x – y) + xy. The term 100 – x – y means to subtract the difference of one number from the other number. That term is multiplied by 100 so it becomes the first two digits. Then the product of the differences, xy, is the last two digits.

Solutions to Practice Problems

95 x 98. The number 95 is 5 less than 100 and the number 98 is 2 less than 100. So we subtract one difference from the other number to get that 95 – 2 = 93 = 98 – 5. So 93 are the first two digits. Then we take the product of the differences, 5 x 2 = 10. The result is 9,310.

96 x 97. The number 96 is 4 less than 100 and the number 97 is 3 less than 100. So we can do 96 – 3 = 93, or we can do 97 – 4 = 93. These are the first two digits. Then the last two digits are equal to the product of the differences, 4 x 3 = 12. The result is 9,312.

98 x 99. The number 98 is 2 less than 100 and the number 99 is 1 less than 100. So we can do 98 – 1 = 97, or we can do 99 – 2 = 97. These are the first two digits. Then the last two digits are 1 x 2 = 02 (remembering this product has to be written as a two-digit answer). The result is 9,702.

91 x 92. The number 91 is 9 less than 100 and the number 92 is 8 less than 100. So we can do 91 – 8 = 83, or we can do 92 – 9 = 83. These are the first two digits. Then the last two digits are 9 x 8 = 72. The result is 8,372.

942. Notice we can write this as 94 x 94. The number 94 is 6 less than 100. So we can do 94 – 6 = 88. These are the first two digits. Then the last two digits are 6 x 6 = 36. The result is 8,836. This problem illustrates the method can be used to square numbers in the 90s easily!



Multiply Two Numbers From 101 To 109

Can you multiply 102 and 107 in your head? There is a similar procedure to the one for multiplying numbers in the 90s. The only difference is the numbers are larger than 100, so we need to add the difference of one number to the other.

There are three steps to the method. First, consider the difference of each number from 100. Second, add the difference of one number to the other number. That result is the first three digits of the answer. Finally, multiply the differences, and that result becomes the last two digits of the answer.

For example, let's do 102 x 107. The number 102 is 2 more than 100, and the number 107 is 7 more than 100.

We add the difference of one number to the other number. So we add 2 to 107 to get 109. We could also add 7 to 102 to get 109. Notice that both results are the same so it doesn't matter which one we do. This result of 109 is the first part of our answer.

Finally, we multiply the differences. So we multiply 2 by 7 to get 14. These are the last two digits of the answer. Putting it together, we have 102 times 107 is equal to 10,914.

Take some time to learn this trick and the one about multiplying numbers in the 90s as the next few tricks will build upon the methods in these tricks.

Practice Problems

105 x 108

106 x 107

108 x 109

101 x 102

1042

Proof

Two numbers in the 100s are 100 + x and 100 + y for x and y less than 10. The product is (100 + x)(100 + y) = 10,000 + 100y + 100x + xy, which is equal to 100(100 + x + y) + xy. The term 100 + x + y is the part about adding one difference to the other number. That is multiplied by 100 so those numbers become the leading digits. Then the product of the differences, xy, is the last two digits.

Solutions to Practice Problems

105 x 108. The number 105 is 5 more than 100 and the number 108 is 8 more than 100. We add the difference of one number to the other number. So we can add 105 + 8 = 113, or we add 108 + 5 = 113. These are the first three digits. Then the last two digits are the product of the differences, 5 x 8 = 40. The result is 11,340.

106 x 107. The number 106 is 6 more than 100 and the number 107 is 7 more than 100. So we can do 106 + 7 = 113, or we can do 107 + 6 = 113. These are the first three digits. Then the last two digits are the product of the differences, 6 x 7 = 42. The result is 11,342.

108 x 109. The number 108 is 8 more than 100 and the number 109 is 9 more than 100. So we can do 108 + 9 = 117, or we can do 109 + 8 = 117. These are the first three digits. Then the last two digits are the product of the differences, 8 x 9 = 72. The result is 11,772.

101 x 102. The number 101 is 1 more than 100 and the number 102 is 2 more than 100. So we can do 101 + 2 = 103, or we can do 102 + 1 = 103. These are the first three digits. Then the last two digits are the product of the differences, 1 x 2 = 02 (remembering to write this as a two-digit number). The result is 10,302.

1042. Notice we can write this as 104 x 104. The number 104 is 4 more than 100. So we can do 104 + 4 = 108. These are the first three digits. Then the last two digits are 4 x 4 = 16. The result is 10,816.



Multiply Two Numbers Near 1,000

Just as we generalized squaring numbers around 50 to the trick of squaring numbers around 500 and 5,000 and so on, we can generalize the trick for multiplying numbers around 100 to higher powers of 10.

The steps are the following. We consider the differences of the two numbers (to 1,000 or 10,000 or etc.). If the two numbers are smaller, we need to subtract the difference of one number from the other. If they are larger then we add the difference of one number to the other. Finally, we multiply the differences and write it with the appropriate number of digits (when the two numbers are smaller, we match the number of digits for the first part of our answer. When the two numbers are larger, we do one fewer digit than the first part of our answer).

For example, let's do 996 times 994. The differences from 1,000 are 4 and 6. Since the numbers are smaller, we will subtract 6 from 996 to get 990 as the first part of the answer. Then we multiply the differences to get 24, which we need to write as a three-digit number 024 (this is because we match the same number of digits in 990, which is a three-digit number). So the answer is 990,024.

Similarly, let's do 1,004 times 1,003. The differences from 1,000 are 4 and 3. We will add 3 to 1,004 to get 1,007 as the first part of the answer. Then we multiply the differences to get 12, which we need to write as a three-digit number 012 (this is because 1,007 has four digits and we need one fewer digit). So the answer is 1,007,012.

Practice Problems

995 x 998

1,006 x 1,007

998 x 999

1,001 x 1,002

1,0042

Proof

Two numbers near 10k, can be written as 10k – x and 10k – y. Note the variables x and y can be positive, for numbers smaller than 10k, or they can be negative, for numbers greater than 10k. We multiply these numbers to get (10k – x)(10k – y) = 102k – 10ky – 10kx + xy, which is equal to 10k(10k – x – y) + xy. The term 10k – x – y is the part about compensating one difference to the other number. That is multiplied by 10k so those numbers become the leading digits. Then the product of the differences, xy, is the last k digits.

Solutions to Practice Problems

995 x 998. The number 995 is 5 less than 1,000 and the number 998 is 2 less than 1,000. So we can do 995 – 2 = 993, or we can do 998 – 5 = 993. These are the leading digits. Then the last three digits are the product of the differences, 5 x 2 = 010 (remembering to write this as a three-digit number). The result is 993,010.

1,006 x 1,007. The number 1,006 is 6 more than 1,000 and the number 1,007 is 7 more than 1,000. So we can do 1,006 + 7 = 1,013, or we can do 1,007 + 6 = 1,013. These are the leading digits. Then the last three digits are the product of the differences, 6 x 7 = 042. The result is 1,013,042.

998 x 999. The number 998 is 2 less than 1,000 and the number 999 is 1 less than 1,000. So we can do 998 – 1 = 997, or we can do 999 – 2 = 997. These are the leading digits. Then the last three digits are the product of the differences, 2 x 1 = 002. The result is 997,002.

1,001 x 1,002. The number 1,001 is 1 more than 1,000 and the number 1,002 is 2 more than 1,000. So we can do 1,001 + 2 = 1,003, or we can do 1,002 + 1 = 1,003. These are the leading digits. Then the last three digits are the product of the differences, 1 x 2 = 002. The result is 1,003,002.

1,0042. Notice we can write this as 1,004 x 1,004. The number 1,004 is 4 more than 1,000. So we can do 1,004 + 4 = 1,008. These are the leading digits. Then the last three digits are 4 x 4 = 016. The result is 1,008,016.



Multiply Two Numbers From 201 To 209, 301 To 309, Etc.

Can you multiply 202 and 207 in your head? We can modify the method to multiply numbers between 101 to 109, and this method will work for other numbers near multiples of 100 like numbers near 200, 300, etc.

There are four steps to the method. First, multiply the leading digits in both numbers. The numbers 202 and 207 have a leading digit of 2, so the first digit in the answer will be 2 x 2 = 4.

Second, consider the difference of each number from the closest multiple of 100. The number 202 is 2 more than 200, and the number 207 is 7 more than 200.

Third, add the differences together and multiply that by the leading digit. So we add 2 and 7 to get 9, and then multiply by the leading digit 2 to get 18. These are next two digits of the answer.

Fourth, multiply the differences of 2 and 7 to get 14. These are the last two digits.

Putting it all together, we have 202 x 207 = 41,814.

Let's do another example of 302 x 301. We multiply the leading digits to get 3 x 3 = 9.

Next, we consider the differences of 2 and 1, and then add them to get 3. We multiply by the leading digit of 3 to get 9. The result from this step is supposed to be a two-digit answer. So we write 9 as 09.

Finally, we multiply the differences to get 2 x 1 = 2. Again we write this as a two-digit answer so the final two digits are 02.

Putting it together, we have 302 x 301 = 90,902.

Let's do a third example which illustrates one more special case. Let's do 709 x 707. We multiply the leading digits to get 7 x 7 = 49. (Since the leading digits are always the same, this is the same as squaring the leading digit.)

Next, we add the differences of 9 and 7, which is 16. We multiply by 7 to get 112. The result from this step is supposed to be a two-digit answer. So we have to carry over the hundreds digit of 1 to the previous part of the answer. In other words, we add the value of the hundreds digit, 1, to the result from the last step, 49, to get 50. These are the leading digits of the answer. Then the two digits 12 are the next two digits of the answer. So the answer is something like 5,012 at this point.

Finally, we multiply the differences to get 9 x 7 = 63. These are the last two digits of the answer.

Putting it together, we have 709 x 707 = 501,263.

Here is the procedure in general to multiply numbers just above 200, 300, etc. Let Y denote the leading digit of the two numbers, so the two numbers are Y0A and Y0B, for A and B between 1 and 9.

1. Multiply Y by itself, which is Y2. The leading digit or digits of the answer will be the number Y2 .

2. Consider the difference of each number to the closest multiple of 100. This will be the numbers A and B.

3. Add the differences and multiply by Y. The next digits of the answer are Y(A + B). Always write this as a two-digit number, so 6 would be written as 06. If the answer has three digits, then carry over the leading digit (in the hundreds spot) to the value of Y2 from the previous step.

4. Finally, multiply the differences. The last two digits are A times B. Again, always write this as a two-digit number, so 6 would be written as 06.

Let's do a final example to illustrate the process. Let's do 909 x 908. The leading digit is Y = 9 and the differences are A = 9 and B = 8.

From the first step, we calculate Y2 = 81.

Then we add the differences and multiply by the leading digit Y. So we have 9 x (9 + 8) = 9 x 17 = 153. This is a three-digit result, so we add the hundreds digit of 1 to the value of Y2 = 81. Thus our answer starts out as 82 and then the next two digits are 53.

Finally, we multiply the differences to get 9 x 8 = 72. These are the last two digits.

Therefore, 909 x 908 = 825,372.

Practice Problems

205 x 208

306 x 307

408 x 409

601 x 602

7042

Proof

Two numbers above a multiple of 100 can be written as 100Y + A and 100Y + B. The product is equal to (100Y + A)(100Y + B), which is 10,000Y2 + 100YB + 100YA + AB = 10,000Y2 + 100Y(A + B) + AB. The term 10,000Y2 refers to squaring the leading digit. Then the next term 100Y(A + B) means to multiply A plus B by the leading digit Y, and then shift it over two places because it is multiplied by 100. Finally, the term AB is the product of the differences, which becomes the last two digits.

Solutions to Practice Problems

205 x 208. The number 205 has a difference of 5 from 200 and the number 208 has a difference of 8 from 200. First we square the lea