MENTAL MULTIPLICATION (FOR SOCIAL SCIENTISTS) MADE EASY BEING A VERY BASIC INTRODUCTION FOR SOCIAL SCIENTISTS AND THE LIKE TO THOSE AMAZING METHODS OF SEEMINGLY MIRACULOUS RAPID MENTAL RECKONING WHICH ARE COLLECTIVELY CALLED BY THE TERRIFYING NAME OF “SPEED MULTIPLICATION.”
J. T. Manhire* ABSTRACT This outline provides a basic introduction to techniques used to perform mental multiplication—including squares—quickly and accurately. It is intended for social scientists and other non-S.T.E.M. professionals obligated to deal with numbers on a daily basis (e.g., tax attorneys, accountants, etc.). After reviewing this brief paper and practicing the techniques, readers should be able to perform complex multiplication calculations in a matter of seconds.
CONTENTS I. II. III.
INTRODUCTION ........................................................... xxx A BRIEF NOTE ON RAPID ADDITION ........................... xxx MULTIPLICATION ........................................................ xxx A. Multiplying by 11 ................................................. xxx B. Complementary Multiplication (Base Methods) ..xxx C. Star Method.......................................................... xxx D. Multiplying Any Pair of Two-Digit Numbers ...... xxx E. Squares.................................................................xxx CONCLUSION .............................................................. xxx
IV.
*
[email protected]. The subtitle is a paraphrase of Silvanus Thompson’s classic book. See SILVANUS P. THOMPSON, CALCULUS MADE EASY: BEING A VERYSIMPLEST INTRODUCTION TO THOSE BEAUTIFUL METHODS OF RECKONING WHICH ARE GENERALLY CALLED BY THE TERRIFYING NAMES OF THE DIFFERENTIAL CALCULUS AND THE INTEGRAL CALCULUS (1910). “S.T.E.M.” is a common acronym standing for Science, Technology, Engineering, and Mathematics. “Math” and “Maths” are synonymous throughout. Nothing herein represents the positions or policies of the government of the United States or any of its agencies. All analyses, conclusions, and mistakes belong solely to the author.
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I. INTRODUCTION Quickly, what is the 𝑟 2 value (coefficient of determination) when the correlation coefficient 𝑟 is 0.45? You have 103 survey respondents and each survey has 107 questions. How many total survey answers do you need to analyze? Your client can take an additional tax deduction of $72 per dependent and she is claiming 11 dependents. How much is her total additional deduction? Oh yeah, one more thing: make sure you get each answer in less than five seconds without a calculator or scratch pad. In other words, do the calculations in your head. To those of you now belly-laughing in disbelief at the thought that you could actually perform such operations in less than five seconds, and completely in your head, I assure you that after reading this brief monograph and practicing its very simple techniques, these three calculations will be child’s play for you. You will not only be more productive in your profession, but you will amaze your friends and terrify your enemies with your seemingly supernatural powers of mental calculation. I know most of you are terrified of math. You’re not alone. We live in a largely “mathphobic” world and most of us do whatever we can to avoid dealing with calculations…especially mental calculations.1 If fact, I’ll bet that if you’re reading this you probably thought twice about even downloading this paper because it had “multiplication” in the title, and the only reason you got this far is because I’ve been able to keep you mildly curious. If you consider yourself bad at math, it’s not your fault. Like most of the western world, you were taught from a very young age to do calculations in the most difficult and arduous way possible. If I believed in conspiracies, I’d say there must be a conspiracy of ignorance in institutionalized education. How else can these terrible techniques be 1 The technical term for the fear of math is “arithmophobia,” which generally means the fear of numbers, but also includes the fear of mathematical calculations. See, e.g., Paula J. Williams, Kris Anne Tobin, Eric Franklin, and Robert J. Rhee, Tackling “Arithmophobia”: Teaching How to Read, Understand, and Analyze Financial Statements, 14 TENN. J. BUS. L. 341, 341 (2013).
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taught generation after generation? Of course, there is no conspiracy. The truth is, most math teachers only know the old and difficult methods because that’s how they were taught and it’s what the textbook tells them to teach, so they teach those methods and the cycle perpetuates itself. Being afraid of math is not shameful, but it is curable. That’s what I’m hoping to accomplish with this paper. I want to cure as many mathphobes as possible by showing you just how easy math can be. You do not need to be born with special abilities or be some sort of “rain man” to excel at mental multiplication. These are simple techniques that anyone can learn very quickly. Let me prove to you how easy mathematics can be, beginning with the 𝑟 2 problem. When squaring any number ending in the digit 5, the last two digits of the answer will always be 25. To get the first two digits, simply add 1 to 4 (the first digit of the number to be squared) to get 5, and then multiply the two numbers (4 × 5 = 20). Put the first and last two digits together and you get your initial answer: 2025. Since you’re squaring a number with two digits after the decimal place, move the decimal of your initial answer four places to the left and you get your final answer: 𝒓𝟐 = 0.2025. If you have 103 respondents to a survey with 107 questions, you want to find the answer to 103 × 107. To get the first part of your answer, add the last digit of the second number (7) to the first number (103). This gives you 103 + 7 = 110. Since 103 and 107 are close to 100 (which has two zeroes), you need two more digits to get your answer (1 1 0 _ _). Take the difference of the first number and 100 (103 − 100 = 3) and the same for the second number (107 − 100 = 7). Multiply 7 and 3 to get 21 and those are the last two digits of your answer (1 1 0 2 1). So your answer is 11,021 questions to analyze. The client with 11 dependents and a $72 deduction per dependent is even easier. For 72 × 11, simply split the numbers 7 and 2 and make room for a third number in between (7 _ 2). Add 7 and 2 to get 9 and that becomes your third digit (7 9 2). So your answer is an additional deduction of $792. With just a little practice, you’ll easily be able to perform these and much more (seemingly) difficult math operations in your head in a
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matter of seconds. This outline provides a very basic introduction to techniques used to perform mental mathematics quickly and accurately. It is intended for social scientists and other professionals obligated to deal daily with numbers (e.g., tax attorneys, accountants, etc.). Mental multiplication is based on just two foundational concepts. The first is a base. This paper refers to standard bases, such as base 10 and base 100, throughout (like the survey problem). Another general concept is splitting. This means that you physically split the answer into at least two separate answers and then combine them to get the final answer (like the tax deduction and 𝑟 2 problems). Familiarity with these foundational concepts will help you succeed with these techniques, but only practice can ensure success. Like anything else worth doing, you must commit to practicing these new concepts daily…even if only for five to ten minutes. This paper only addresses mental multiplication, with a brief mention of rapid addition. Other common operations, such as subtraction, division, fractions, cubes, and roots, are not included. These will be covered in future work. Of course, this is only a brief introduction, and there are many more techniques that lie below the surface of this proverbial iceberg.2 The concepts and techniques are fundamental, and any practitioner of this craft will want to pursue this discipline deeper and broader than what is exposited here. Armed with this short paper alone, anyone can become a mental mathematics tyrannosaurus. I ask only that you use your newly-found powers for good instead of evil.
2 Some of the first popular books on “speed math” include ANN CUTLER & RUDOLPH MCSHANE, THE TRACHTENBERG SPEED SYSTEM OF BASIC MATHEMATICS (1960); HENRY STICKER, HOW TO CALCULATE QUICKLY: FULL COURSE IN SPEED ARITHMETIC (1945); EDWARD STODDARD, SPEED MATHEMATICS SIMPLIFIED (1965); GERARD W. KELLY, SHORT-CUT MATH (1969). Other popular books that explore mental mathematics in greater detail include EDWARD H. JULIUS, RAPID MATH TRICKS & TIPS (1992); EDWARD H. JULIUS, ARITHMETRICKS: 50 EASY WAYS TO ADD, SUBTRACT, MULTIPLY, AND DIVIDE WITHOUT A CALCULATOR (1995); BILL HANDLEY, SPEED MATHEMATICS: SECRET SKILLS FOR QUICK CALCULATION (2003); SCOTT FLANSBURG, MATH MAGIC: HOW TO MASTER EVERYDAY MATH PROBLEMS; DHAVAL BATHIA, VEDIC MATHEMATICS MADE EASY (2005); ARTHUR BENJAMIN & MICHAEL SHERMER, SECRETS OF MENTAL MATH: THE MATHEMAGICIAN’S GUIDE TO LIGHTNING CALCULATION AND AMAZING MATH TRICKS (2006);
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II. A BRIEF NOTE ON RAPID ADDITION Although I am purposely restricting the scope of this paper to multiplication, many of the multiplication techniques require some rapid addition. Therefore, I will spend just a few paragraphs discussing how to do “speed addition.” Most of us know how to add from right to left. Start with the ones column, then the tens column, hundreds column, etc. Most mental math techniques involve adding in the opposite direction; from left to right. The trick is to remember which column you are in at any one time (hundreds, tens, ones, etc.). For example, take a look at the equation:
+
35 67 56
Most of us can do this in our head if we really try, but we usually go about it by adding the ones column (5 + 7 + 6 = 18; bring down the 8; carry the one), and then we add the tens column (3 + 6 + 5 = 14 plus the carried 1 = 15). Bring down the 15 and we have the answer: 158. This method works, but it very slow and difficult to do mentally. Instead, try adding from left to right starting with the tens column. Recall that it’s called the tens column because all of the digits in the column have the value of the digit times 10. So the digit 3 in 35 has the value 30 (3 × 10), the digit 6 in 67 has the value 60 (6 × 10), and so on. So when you see the 3 in the tens column, think “30;” a 6, think “60;” a 5, think “50.” You get the picture. So for the equation 35 + 67 + 56, first think “30 + 60 + 50.” Most of us can do this pretty quickly and get to 140. Then just keep counting in the ones column: 140 + 5 = 145; 145 + 7 = 152; and 152 + 6 = 𝟏𝟓𝟖. We get to the same answer, just faster.
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Try another one:
+
26 11 79
Go to the tens column and start adding up the tens; “20 + 10 + 70 = 100.” Then continue with the ones column; “100 + 6 = 106; 106 + 1 = 107; 107 + 9 = 116. And you’re done. You can even start looking for groupings that add to 5 or 10 to make it easier. For example, once you add up the tens column and get 100, you can quickly scan the ones column and see that there’s a grouping of 9 and 1 to quickly get to 110. After that, you just have to add the 6 to get to 116. See, this stuff is actually pretty easy. Now let’s try adding three digit numbers. Take 326 678 245 + 567
326 + 678 + 245 + 567. Use the same technique only this time start with the hundreds column, which means every digit is multiplied by 100. Start from the left and go to the right: 300 + 600 + 200 + 500 = 1600. Then keep adding with the tens columns: 1600 + 20 + 70 + 40 + 60 = 1790. Then go to the ones column: 1790 + 6 + 8 + 5 + 7 = 1816. And there’s your answer. Although I’m using equations for this paper, I do not think with plus signs, and maybe you don’t either. What works for me is simply counting off the sums as I go. So instead of 300 + 600 + 200 + 500 = 1600, my brain goes “300, 900, 1100, 1600.” Same with the tens and ones. Everyone is a little different, so practice, practice, practice and find what works best for you.
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III. MENTAL MULTIPLICATION Rapid addition can be mildly amusing, but the truly amazing mental math starts with multiplication. It’s amazing because most of us think multiplication is so difficult, so when someone calculates the product of 98 and 96 in less than two seconds, our jaws drop because inside we are thinking, “there’s no way I can do that.” But you can. In fact, some of the multiplication techniques are easier than the addition technique. A. Multiplying by 11 I want to start with something simple so you can quickly amaze yourself. Why? Because success breeds success, and not only hitting the ball but hitting it hard the very first time you step up to the plate will give you the confidence you need to continue reading, which means you will continue learning. So let’s start with a quick and easy way to multiply any number (even a big one) by 11. I will start with some special techniques and end with a general technique that applies to any number you want to multiply by 11 quickly without ever swiping for that calculator app. 1. Single Digits: Clone Them The special technique for single digits is one you probably already know. All you do is clone the single digit by which you are multiplying 11. For example, if you multiply 11 by 1, you simply take the number 1 and create a clone (another 1) and put them next to each other. So the problem would read 11 × 1 = 11. Notice how the answer is simply a clone of the original number (1 in this case) placed next to the original. The same is true for 11 × 2. Since 2 is the original number, clone it (make another 2), and place it next to the original. So the equation reads 11 × 2 = 22. You can repeat this pattern for all single digit numbers, including zero.
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Here are the rest: 11 × 3 = 33 11 × 4 = 44 11 × 5 = 55 11 × 6 = 66 11 × 7 = 77 11 × 8 = 88 11 × 9 = 99 and 11 × 0 = 00. 2. Double Digits: Split and Add I realize most of you already knew the cloning technique for single digits, but I’ll bet this next technique will be new to most of you. For any two-digit number (10 through 99), simply split the first and second numbers apart, and then put the sum of the original two digits in the middle. For example, if you are multiplying 27 by 11, take the number 27 and split the number into two “bookends” so you have the 2 on the left and the 7 on the right with room for one more number in between them, like this: 27 → 2 ___ 7. After you split, add the original two digits together and put their sum in the middle of the split numbers. Since the sum of 2 and 7 is 9, put the 9 in the middle of 2 and 7 like this: 27 → 2 __ 7 2+7=9→2 𝟗 7 and you have your answer: 11 × 27 = 297. Go ahead…double check it on your calculator app. Let’s do another one. Try 11 × 78. Step 1, split 78 into two bookends so you have 7 on the left and 8 on the right, with room for a third number in the middle, like this:
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78 → 7 __ 8. Step 2, add 7 and 8 and place the sum in between the bookends, like this: 78 → 7 __ 8 7 + 8 = 15 → 7 𝟏𝟓 8. Notice that the sum gives you the two-digit number 15, but when you split 7 and 8 you only have room for one digit. When this happens you need to carry the number in the tens place (1 is the number in the tens place for the number 15) and add it to your left bookend number (7 in this case) and keep only the ones digit of the sum (5 in this case). So with the result 7 15 8, you carry the 1 and add it to the 7 and leave the 5 in the middle. Since 7 + 1 = 8, your new left bookend number is 8, so 7 15 8 becomes 8 5 8, and that’s your answer. 11 × 78 = 858. What does your calculator app say? It’s right, isn’t it? 3. Three-Digits and More: Add to the Neighbor When you multiply digits greater than 99 by 11, you need to use a general technique. This general technique is sometimes called adding to the neighbor. It simply means that you add a digit to the digit immediately to the right. Let’s look at some examples. Say you want to mentally multiply the number 12,345 by 11. First, put lines to the left and right of the number and add zeros on either side (when first trying this, use a paper and pencil, but pretty quickly you’ll be able to set it up in your head). Your new number should look like this: 0 | 12,345 | 0. Next, add to the neighbor. Start with the last digit of the original number (5 in this case) and add it to the right-most zero. 5+0=𝟓
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This will be the right-most digit of your answer. Then add the neighbor to the left of the last digit of the original number and the last digit of the original number. So in this case the neighbors would be 4 and 5. 4+5=𝟗 Keep adding neighbors, moving all the while to the left, until you add the two left-most digits (0 and 1 in this case), like this: 3+4=𝟕 2+3=𝟓 1+2=𝟑 0+1=𝟏 If you line up all of your sums from right to left, you get your answer. So line the boldfaced digits up from right to left and you get: 1 3 5 7 9 5. And that’s your answer. 11 × 12,345 = 𝟏𝟑𝟓, 𝟕𝟗𝟓. Let’s do another one, this time where we have to carry a digit. Multiply 5,179 by 11. Again, set up the number with zero bookends like this: 0 | 51790 | 0 Next, add the neighbors from the right starting with 9 + 0 = 9, which gives you the last digit of your answer. When you add the next neighbors, 7 and 9, notice that 7 + 9 = 16. Here, the 6 becomes the second digit of your answer and you need to carry the 1 over to the next round of adding. This means when you add the next neighbors, 1 and 7, you need to also add the 1 you carried over from the 16. So the third digit of your answer will be 1 + 7 + (1) = 9. Next, add the 5 and 1 to get 5 + 1 = 6, and finally add the 0 and 5 to get the last digit of your answer, 0 + 5 = 5. Here’s a recap of all the “add to the neighbors” from the problem. The bold digits are the ones that become your answer. Notice that you must carry the 1 in the second operation:
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9+0=𝟗 7+9= 1𝟔 1 + 7 + (1) = 𝟗 5+1=𝟔 0+5=𝟓 So your answer is 5,179 × 11 = 𝟓𝟔, 𝟗𝟔𝟗. 4. An Alternate Method There is an alternate method for multiplying any number by 11. Its general form might seem scary because it looks a lot like algebra, but it’s actually quite simple. The formula is 11 × 𝑛 = 𝑛 + 10𝑛. Let’s walk through it together. All you do is take the number by which you’re multiplying 11 and call that 𝑛. So in the last two-digit example 11 × 78, 𝑛 = 78. Using our alternate formula and plugging 78 in for 𝑛 we get 11 × 78 = 78 + (10 × 78). Since 10 × 78 = 780, we find that 11 × 78 = 78 + 780. Adding from left to right we get 700, 850, 858 (see the addition section for a review if necessary). And that’s our answer: 858. This method works for any value of 𝑛, so it’s pretty versatile and some might find it easier than the “add to your neighbor” method. Again, practice, practice, practice and play around with the different techniques until you find what works for you. B. Complementary Multiplication (Base Methods) As mentioned in the introduction, base methods—also known generally as complementary multiplication—are some of the most popular in mental mathematics, and for good reason. Not only are they efficient methods for mental calculation, but they also cause the greatest amazement in those observing someone using the technique. We are trained from a young age that multiplying small numbers is easier than multiplying large numbers. The base methods introduced here will show you that this is not necessarily true. Large numbers can be multiplied just as easily as small numbers.
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Although the next section covers the vertical and crosswise method of multiplying any two numbers, the techniques themselves will be introduced with the base methods. A base is just a “home number;” one you can remember easily and one that is easy to use in calculations. The most popular bases with this method are 10 and 100, but other bases can and should be used, such as 20, 50, 200, and even 1,000. There are really only two rules for the base methods that you need to remember: (1) add across diagonally, and (2) multiply vertically (up and down). Let’s see this simple rule in action. Let’s start with a pair of numbers that are each less than 10. This is the best introduction since you probably know your multiplication tables from 1 to 10 from rote memory. Notice I didn’t say you know how to calculate these numbers. You only know them through memorization. Let’s look at how to actually calculate them, starting with 7 × 6. Of course you know the answer is 42 from your multiplication tables. That’s why this is a good introduction to the method…you already know the answer. First, figure out the difference between each number and the base 10. Since 10 − 3 = 7, we know that 7 is −3 from the base 10. Likewise, since 10 − 4 = 6, we know that 6 is −4 from the base 10. With this information we can set up the problem with the differences off to the right side (do this on paper for now, but soon you will be able to set it up in your head).
×
7 6
−3 −4
First, add crosswise, so either add 7 and −4 or 6 and −3. As you can see it doesn’t matter which one you choose because you always get the same answer. In this case the answer is 3. Place the 3 in the left bottom box under the original equation like this:
×
7 6 3
−3 −4
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Now multiply vertically (up and down) the numbers in the rightmost column. Multiplying −3 × −4 = 12. The 12 goes in the right bottom box, but only one digit will fit there. So put the 2 in the right bottom box and carry the 1 over and add it to the 3. This will give you 4 in the left bottom box and 2 in the right bottom box, like this:
×
7 6 3𝟒
−3 −4 𝟐
Put the numbers together and you have your answer. 7 × 6 = 𝟒𝟐. And you’re done. Let’s do one more and then we’ll step it up. Let’s do 9 × 8, which you already know is 72. Set up the equation like before.
×
9 8
−1 −2
The −1 is next to the 9 since that is the distance from the base 10. It’s the same for the −2 and the 8. Next, add crosswise. Remember you can add 9 and −2 or 8 and −1. They both give you the same result, which is 7. Place the 7 in the bottom left box.
×
9 8 7
−1 −2
Now multiply vertically on the right. Since −1 × −2 = 2, put the 2 in the bottom right box.
×
9 8 𝟕
−1 −2 𝟐
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Combine the numbers and you have your answer: 9 × 8 = 72. Let’s do the same thing but instead of 10, let’s use 100 as our base. The equation is 98 × 97. The major difference between using base 10 and base 100 is that the bottom right box only has room for one digit with base 10, but the box must have two digits with base 100. The first step is to figure out how far away each number is from your base. 98 is −2 from the base 100, and 97 is −3 away. The next step is to set up your table (on paper or in your mind) like before, only using two digit numbers instead of one.
×
98 −02 97 −03
The next step is to add crosswise (you can pick), which gives you 95. Put the 95 in the bottom left box.
×
98 −02 97 −03 𝟗𝟓
Lastly, multiply vertically. Since −02 × −03 = 06, put the 06 in the bottom right box.
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98 −02 97 −03 𝟗𝟓 𝟎𝟔
And you’re done. 98 × 97 = 𝟗, 𝟓𝟎𝟔. You probably notice that all of our examples so far dealt with numbers below the base. Let’s look at a few equations with numbers above the base. We’ll start again with base 10 and solve the equation 12 × 13. Like before, we need to determine how far each number if from the base, but since the numbers are above the base the differences
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will be positive instead of negative. 12 is 2 away from base 10 and 13 is 3 away. Next, set up your table as before, but this time remember that the differences on the right are positive and not negative.
×
12 13
+2 +3
Not add crosswise. Bothe 12 + 3 and 13 + 2 give you 15, so put the 15 in the bottom left box.
×
12 13 15
+2 +3
Multiply the differences and you’ll get 2 × 3 = 6. Place the 6 in the bottom right box. Since we’re only in base 10, the bottom right box can only have one digit.
×
12 13 𝟏𝟓
+2 +3 𝟔
And we’re done. 12 × 13 = 𝟏𝟓𝟔. Now let’s multiply two numbers that are above base 100. We’ll try103 × 107. The difference between 103 and 100 is 3 and the difference between 107 and 100 is 7. Now you can set up your table, but remember the bottom right box must have two digits since the base is 100. 103 +03 × 107 +07
Add crosswise to get 110, which you place in the bottom left box.
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103 +03 × 107 +07 110 Multiply the differences vertically to get 03 × 07 = 21. Place the 21 in the lower right box and we’re done. 103 +03 × 107 +07 𝟏𝟏𝟎 𝟐𝟏 103 × 107 = 𝟏𝟏, 𝟎𝟐𝟏. If you use base 100 and the number in the bottom right box has more than three digits, put the two right-most digits in the box and carry the left-most digit over to be added to the number in the bottom left box. What is one number is above and the other below the base? Then what? Well, it’s pretty much the same method. It is probably best to explain with an example. Let’s go to base 10 and solve 13 × 9. 13 is 3 away from 10 and 9 is −1 away from 10. Like before, set up your table and add crosswise. 13 − 1 = 12 and 9 + 3 = 12, so 12 goes in the bottom left box. +3 × −1 = −3, so −3 goes in the bottom right box.
×
13 9 12
+3 −1 −3
Here’s where it gets different from a situation when both numbers are above or below the base. You clearly can’t just combine 12 and −3, so what we do is multiply the bottom left number by the base (12 × 10 = 120) and add the negative 3 (120 − 3 = 117). And now we’re done. 13 × 9 = 𝟏𝟏𝟕. You do the same thing when multiplying one number above and one number below base 100, except in the last step you would multiply the bottom left number by 100 instead of 10.
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Now what if your numbers are not close to base 10 or 100. Then what? Again, it’s pretty much the same idea, except you need to do one more step if you base is not a 10 or 100 (or 1,000; 10,000; etc.). Again, let’s take an example and work through it. Let’s solve 23 × 21. Both numbers are very close to 20, so 20 we’ll use 20 as our base. Like before, find the difference between the numbers and the base. These numbers are both above the base so the differences will be positive. 23 is 3 more than base 20, and 21 is 1 more. Now set up the table, add crosswise, and multiply vertically.
×
23 21 24
+3 +1 3
Here’s the difference since we are in base 20. You have to multiply the bottom left number by the tens digit of the base and then combine it with the bottom right number. So in this case the bottom left number is 24 and the tens digit of the base is 2. 24 × 2 = 48. This becomes your new bottom left number.
×
23 21 24 𝟒𝟖
+3 +1 3 𝟑
Combine the new bottom left number with the bottom right number and you have your answer. 23 × 21 = 𝟒𝟖𝟑. You might wonder why there is an extra step for base 20 than for base 10. Actually, the steps are exactly the same. Before combining the bottom boxes, you need to always multiply the left bottom number by the tens digit of the base. It’s just that the tens digit of the base in base 10 is 1, so multiplying the bottom left number by 1 just gives you the same number you started with. For simplicity, we just leave that step off for base 10, but remember that technically it’s still there. Of course, if the base was 30 you’d multiply the bottom left number by 3, if the base is 40, multiply by 4, etc.
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Try doing some base method multiplication before moving on to the next section. When starting out, work on a sheet of paper or a whiteboard, but after a few exercises, try to do the steps in your head. Remember the steps:
Find your closest base Find how far each number is from the base; positive if above the base, negative if below Make your table (on paper or in your mind) Add crosswise to get your bottom left number Multiply vertically to get your lower right number Multiply the lower left number by the left-most digit of the base to get your new bottom left number (skip this step if the leftmost digit is 1) Combine the bottom left and right numbers, and you’re done!
Once you get used to it, you’ll be able to very quickly multiply seemingly large numbers at record speed. C. Star Method The star method appears more difficult at first than it actually is. Once you understand the star pattern and practice a little, you’ll be multiplying four- and five-digit numbers in a snap. Let’s use an example to explain the method. Solve 123 × 456. Set the problem up as you normally would, like this:
×
123 640
Step 1: Multiply the 3 and the 0 to get 0.
×
123 640 ∗∗ 0
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Step 2: Cross multiply the 0 and the 2, cross multiply the 4 and the 3, and add the two products together to get (0 × 2) + (4 × 3) = 12. Put the 2 as the next digit of your answer and carry the 1.
×
123 640 ∗ 21 0
Step 3: Cross multiply the 6 and the 3, cross multiply the 1 and the 0, and vertically multiply the 2 and the 4 (if you draw it out it makes a star pattern). Add the three products together plus the carried 1 to get (6 × 3) + (1 × 0) + (2 × 4) + 1 = 27. Put the 7 as the next digit of your answer and carry the 2.
×
123 640 72 20
Step 4: Cross multiply the 6 and the 2, cross multiply the 4 and the 1, and add the two products together plus the carried 2 to get (6 × 2) + (4 × 1) + 2 = 18. Put the 8 as the next digit of your answer and carry the 1.
×
123 640 81 720
Last Step: Vertically multiply the 6 and the 1, and add the carried 1 to get (6 × 1) + 1 = 7. Put the 7 as the final digit of your answer.
×
123 640 78720
And you’re done. The answer is 123 × 456 = 𝟕𝟖, 𝟕𝟐𝟎. I know this looks complicated, but once you understand the star pattern and practice
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it (this one really does take practice), you start to careen through complicated multiplication problems. Here’s the most common mistake I’ve seen with those starting out: you forget to carry. You need to practice your mental carrying for this method to produce accurate results. I recommend working through a number of problems on paper until you get the method and carrying down. D. Multiplying Any Pair of Two-Digit Numbers This approach is extremely simple, but many people don’t bother learning it because it requires memorizing what looks like an ugly algorithm. However, if you take the time to memorize it, the formula provides a very quick mental multiplication technique. Here’s the formula defining 𝑎 as the tens digit of the first number, 𝑏 as the ones digit of the first number, 𝑐 as the tens digit of the second number, and 𝑑 as the ones digit of the second number: 〈𝑎𝑏〉 × 〈𝑐𝑑〉 = 100𝑎𝑐 + 10(𝑏𝑐 + 𝑎𝑑) + 𝑏𝑑. For example: 37 × 62 = 100(3 × 6) + 10[(7 × 6) + (3 × 2)] + (7 × 2). This reduces to: 1800 + 480 + 14 = 𝟐, 𝟐𝟗𝟒. Again, it takes a little practice to memorize the formula and get comfortable using it, but once you do this method can be a less cumbersome alternative to the vertical and cross method for two-digit multiplication.3 3 There are other techniques for multiplying a pair of two-digit numbers that are based on more formal derivations of quadratics equations. For example, recall from high school algebra that (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 . If 𝑎 is the tens digit and 𝑏 the ones digit, one can imagine the equation 64 × 56 thereby defining 𝑎 as 60 and 𝑏 as 4. Using the quadratic factors yields a solution to the multiplication problem since 64 × 56 = (60 + 4)(60 − 4) = (602 − 42 ) = 3600 − 16 = 3,584. Again, these formal expressions can be helpful if you wish to spend the time learning and practicing them.
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E. Squares Any number squared is simply an operation of multiplying a number by itself. For this reason, it’s important to include techniques for mentally squaring any number in a discussion on rapid mental multiplication. It is probably obvious that since squaring a number is a multiplication operation at its heart, we can use the techniques we’ve already learned to solve any equation involving a square. For example, if we want to solve the equation 112 , we can simply imagine the equation as 11 × 11 and apply any of the methods we’re already familiar with. Since it involves multiplying by 11, we can split the number (1 __ 1), add the integers (1 + 1 = 2) and place the sum in the middle of the split numbers (1 2 1). We’d arrive at the correct answer of 112 = 𝟏𝟐𝟏. Since 11 is close to base 10, we can also use our base method by finding the distance of 11 from base 10 (+1), adding 11 to +1 to get 12, placing the 12 in the bottom left box, and multiplying +1 × +1 = 1 to find the number that goes in the bottom right box. Of course, we get the same answer, 121. And lastly we can use the star method to multiply the right digits vertically (1 × 1 = 𝟏), multiply crosswise and add the products ((1 × 1) + (1 × 1) = 𝟐), and multiply the left digits vertically (1 × 1 = 𝟏). Putting them all together gives us the same answer: 121. But squares are a special form of multiplication, and as such there are a few special tricks that allow us to solve for them faster than with our now-familiar mental multiplication tricks. We’ll first discuss solving for any square ending in the digit “5,” then we’ll discuss the method for solving any square in just seconds. 1. Squares Ending in “5” Recall the 𝑟 2 problem from the introduction. We used the percentage 0.45 for 𝑟, but to make this example simpler, let’s now
Since such advanced methods are beyond the scope of this brief introduction to mental multiplication, this paper does not go into these formulae further.
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simply solve for the square of 45. Here’s something you need to remember: All squares with the last digit “5” will have an answer ending with the last two digits “25.” This is true for all squares ending in “5.” Therefore, we know that 452 = ___ 2 5. To get the first part of the answer, all we do is add 1 to the digit in the ten’s place and multiply the answer times the number in the ten’s place. So in this example since 4 is in the ten’s place, we add 1 to it to get 4 + 1 = 5. We then take the 5 and multiply it by the 4 to get 5 × 4 = 20. We then use the 20 as the first part of our answer (2 0 2 5), and we’re done. The answer is 452 = 𝟐, 𝟎𝟐𝟓. If you’re not sure, check the calculator app. Let’s try another one: 1252 . We know the answer will have the last two digits 25, so you can set up the answer in your mind like this: ____ 2 5. The remaining number in 125 is 12. Adding 1 to 12 gives us 13. Multiplying them together gives us 13 × 12 = 156 (if necessary, use the base 10 method discussed earlier to find 13 × 12 = 156). Make the 156 the first part of the answer (1 5 6 2 5), and you’re done. 1252 = 𝟏𝟓, 𝟔𝟐𝟓. Yes, it really is that simple. If you want to get faster, try going from left to right. For 1252 , add 1 to the number in front of the 5 and multiply (13 × 12 = 156), then just tack on the 25 to the end of the answer to get 1252 = 𝟏𝟓, 𝟔𝟐𝟓. It’s up to you. Again, I encourage you to practice different approaches and use the method with which you are most comfortable. 2. Base 50 Squares The mental method for finding squares ending in “5” is almost too easy, but be very careful…the method does not work if the number being squared does not end in a 5. For other numbers, we need a different method. In this section we will discuss squaring numbers that are close to the base 50. We’ll first discuss squaring numbers below base 50, and then number above (by the way, 502 = 2,500…you might want to just remember that one for speed). If you square a number below base 50, remember to always work with the number “25.” Consider it your sub-base for these
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problems.4 Working through an example might be helpful here, so let’s start by solving 472 . Remember how we found the distance of the number from the base when we discussed the base method? We’ll use the same idea here. The first step is to figure out how far away from base 50 is 47? Since 50 − 3 = 47, we know that 47 is −3 from 50. Step 2 is to add this distance (−3) to the sub-base 25. So we get 25 − 3 = 22. This becomes the first part of our answer (2 2 __ __). To get the second part of the answer we square the difference of the number from base 50, which is −3 in this case. Now the second part of the answer needs to have two digits, so it is helpful to think of −3 as −03. Since −03 × −03 = 09, just use these two digits as the second part of your answer to get 2 2 0 9 and you’re done. The answer is 472 = 𝟐, 𝟐𝟎𝟗. Now let’s try one above base 50. We still need the sub-base 25, so keep that in mind. The only difference is that the distance from 50 will now be positive instead of negative. All other steps are the same. Let’s solve 562 . The first step is to determine the distance of 56 from base 50. Since 50 + 6 = 56, the difference is +06 (again, I suggest using two digits to remember the answer will require two digits at the end). The next step is to add the difference to sub-base 25, and you get 25 + 06 = 31. This becomes the first part of your answer (3 1 __ __). Next, square the difference to get 06 × 06 = 36, make 36 the last two digits of your answer, and you’re done. The answer is 562 = 𝟑, 𝟏𝟑𝟔. If you got 3,112 as your answer, remember that your are multiplying 6 × 6, not adding 6 + 6. Of course, the farther you get from 50, the larger the second part of your answer will be. Always remember that with the base 50 method for squares, the last part of the answer can only hold two-digit…no less and no more. If you get a second part of the answer that’s less than two digits, you must put a zero in front of it. If you get a second part of the 4
How you remember to use 25 as your sub-base is up to you. I always remember that the sub-base 25 is ½ of base 50. It makes sense to me that the sub-base is ½ of the base. Again, this is just my trick, but please use the method that works best for you.
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answer that’s more than two digits, you need to carry the third digit over to the first part of the answer, as in the following example. Solve 622 . The first step is to determine the distance of 62 from base 50, which is 12. Next, add sub-base 25 and 12 to get 37. This becomes the first part of your answer (3 7 __ __). Next, square the distance 12 to get the second part of your answer. Notice that when you square 12 you get 144, which is a three digit number. If this happens, use the 44 as your last two digits and carry the 1 over to the 37 in the first part of your answer. Since 37 + 1 = 38, the first part of your answer changes from 37 to 38 (3 78 4 4). And that’s your answer; 622 = 𝟑, 𝟖𝟒𝟒. 3. Algorithmic Approaches to Squaring Numbers Again, some of the methods for mental multiplication require learning short algorithms. At first glance, these can overwhelm even a seasoned mental calculator; however, they are surprisingly simple if you take the time to learn them and practice them. This section will present each algorithm and then give a brief example. This section will not cover each algorithm in as much detail as previous section, but at this point you probably have enough mental ammunition to work through the examples quickly. I will use a special notation here. I will use 𝑥 to denote the number being squared. This will be for all numbers so 𝑥 can be more than a one-digit number (e.g., 𝑥 2 = 1572 ). If the algorithm breaks out the digits of the number being squared, I will use the notation 𝑥1 for the right-most digit, 𝑥2 for the digit immediately to the left of the right-most digit, etc. So for the number 6,789, 6 = 𝑥4 , 7 = 𝑥3 , 8 = 𝑥2 , and 9 = 𝑥1 (since 9 is the right-most digit). If there is no subscript attached to 𝑥, assume 𝑥 represents the entire number being squared.
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a. Any Two-Digit Number: Algorithm: (𝑥2 𝑥1 )2 = 100(𝑥2 )2 + 20(𝑥2 × 𝑥1 ) + (𝑥1 )2. Example: 372 = 100(3)2 + 20(3 × 7) + (7)2 = 900 + 420 + 49 = 𝟏, 𝟑𝟔𝟗 b. Numbers 52 to 99: Algorithm: 𝑥 2 = (100 − 𝑥)2 + 100[𝑥 − (100 − 𝑥)]. Example: 932 = (7)2 + 100[93 − (7)] = 49 + 8600 = 𝟖, 𝟔𝟒𝟗 c. Numbers 101 to 148: Algorithm: 𝑥 2 = (𝑥 − 100)2 + 100[𝑥 + (𝑥 − 100)]. Example: 1472 = (47)2 + 100[147 + (47)] = 2,209 + 19,400 = 𝟐𝟏, 𝟔𝟎𝟗 d. Numbers Below Base 1,000: Algorithm: 𝑥 2 = (1,000 − 𝑥)2 + 1,000[𝑥 − (1,000 − 𝑥)]. Example: 9942 = (6)2 + 1,000[994 − (6)] = 36 + 988,000 = 𝟗𝟖𝟖, 𝟎𝟑𝟔 e. Numbers Above Base 1,000: Algorithm: 𝑥 2 = (𝑥 − 1,000)2 + 1,000[𝑥 + (𝑥 − 1,000)]. Example: 1,0072 = 72 + 1000[1007 + 7] = 49 + 1,014,000 = 𝟏, 𝟎𝟏𝟒, 𝟎𝟒𝟗
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f. Numbers Ending in 1: Algorithm: 𝑥 2 = (𝑥 − 1)2 + (2𝑥 − 1). Example: 1472 = (47)2 + 100[147 + (47)] = 2209 + 19400 = 𝟐𝟏, 𝟔𝟎𝟗 g. Numbers Ending in 4: Algorithm: 𝑥 2 = (𝑥 + 1)2 − (2𝑥 + 1). Example: 142 = (14 + 1)2 − (28 + 1) = (225) − (29) = 𝟏𝟗𝟔 h. Numbers Ending in 6: Algorithm: 𝑥 2 = (𝑥 − 1)2 + (2𝑥 − 1). Example: 162 = (16 − 1)2 + (32 − 1) = (225) + (31) = 𝟐𝟓𝟔 i. Numbers Ending in 9: Algorithm: 𝑥 2 = (𝑥 + 1)2 − (2𝑥 + 1). Example: 192 = (19 + 1)2 − (38 + 1) = (400) − (39) = 𝟑𝟔𝟏 There are other methods for squaring numbers. Because the objective of this paper is to provide a short introduction to mental multiplication, I will not include these methods here. I do encourage you to investigate and learn these methods on your own from publications like those cited in the introductory footnotes.
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IV. CONCLUSION I hope this brief outline has, at a minimum, convinced you that mental multiplication is achievable by anyone willing to devote the time necessary to master it. Again, it takes practice to become one of those folks you see on TV beating a calculator. But even if you don’t aspire to be “that guy,” the techniques outlined in this brief introduction should be sufficient for you to at least do multiplication in your head faster than before you started reading it. As a social scientist, tax attorney, CPA, actuary, or any similar profession, just being able to do accurate multiplication in your head faster than before can be a significant benefit. It’s also something that’s just really cool. Good luck, have fun, and keep practicing!
