Secrets of Mental Math by Arthur Benjamin (reading fiction .TXT) 📕

= 99, 8 × 12 = 96, 7 × 13 = 91, 6 × 14 = 84, 5 × 15 = 75, 4 × 16 = 64, and so on. I noticed that the products were getting smaller, and their difference from 100 was 1, 4, 9, 16, 25, 36, … – or 12, 22, 32, 42, 52, 62, … (see table below).

I found this pattern astonishing. Next I tried numbers that add to 26 and got similar results. First I worked out 132 = 169, then computed 12 × 14 = 168, 11 × 15 = 165, 10 × 16 = 160, 9 × 17 = 153, and so on. Just as before, the distances these products were from 169 was 12, 22, 32, 42, and so on (see table below).

There is actually a simple algebraic explanation for this phenomenon (see Why These Tricks Work, this page). At the time, I didn’t know my algebra well enough to prove that this pattern would always occur, but I experimented with enough examples to become convinced of it.

Then I realized that this pattern could help me square numbers more easily. Suppose I wanted to square the number 13. Instead of multiplying 13 × 13,

why not get an approximate answer by using two numbers that are easier to multiply but also add to 26? I chose 10 × 16 = 160. To get the final answer, I just added 32 = 9 (since 10 and 16 are each 3 away from 13). Thus, 132 = 160 + 9 = 169. Neat!

This method is diagrammed as follows:

Now let’s see how this works for another square:

To square 41, subtract 1 to obtain 40 and add 1 to obtain 42. Next multiply 40 × 42. Don’t panic! This is simply a 2-by-1 multiplication problem (specifically, 4 × 42) in disguise. Since 4 × 42 = 168, 40 × 42 = 1680. Almost done! All you have to add is the square of 1 (the number by which you went up and down from 41), giving you 1680 + 1 = 1681.

Can squaring a two-digit number be this easy? Yes, with this method and a little practice, it can. And it works whether you initially round down or round up. For example, let’s examine 772, working it out both by rounding up and by rounding down:

or

In this instance the advantage of rounding up is that you are virtually done as soon as you have completed the multiplication problem because it is simple to add 9 to a number ending in 0!

In fact, for all two-digit squares, I always round up or down to the nearest multiple of 10. So if the number to be squared ends in 6, 7, 8, or 9, round up, and if the number to be squared ends in 1, 2, 3, or 4, round down. (If the number ends in 5, you do both!) With this strategy you will add only the numbers 1, 4, 9, 16, or 25 to your first calculation.

Let’s try another problem. Calculate 562 in your head before looking at how we did it, below:

Squaring numbers that end in 5 is even easier. Since you will always round up and down by 5, the numbers to be multiplied will both be multiples of 10. Hence, the multiplication and the addition are especially simple. We have worked out 852 and 352, below:

As you saw in Chapter 0, when you are squaring a number that ends in 5, rounding up and down allows you to blurt out the first part of the answer immediately and then finish it with 25. For example, if you want to compute 752, rounding up to 80 and down to 70 will give you “Fifty-six hundred and … twenty-five!”

For numbers ending in 5, you should have no trouble beating someone with a calculator, and with a little practice with the other squares, it won’t be long before you can beat the calculator with any two-digit square number. Even large numbers are not to be feared. You can ask someone to give you a really big two-digit number, something in the high 90s, and it will sound as though you’ve chosen an impossible problem to compute. But, in fact, these are even easier because they allow you to round up to 100.

Let’s say your audience gives you 962. Try it yourself, and then check how we did it.

Wasn’t that easy? You should have rounded up by 4 to 100 and down by 4 to 92, and then multiplied 100 × 92 to get 9200. At this point you can say out loud, “Ninety-two hundred,” and then finish up with “sixteen” and enjoy the applause!

Computer the following:

Zerah Colburn: Entertaining Calculations

One of the first lightning calculators to capitalize on his talent was Zerah Colburn (1804–1839), an American farmer’s son from Vermont who learned the multiplication tables to 100 before he could even read or write. By the age of six, young Zerah’s father took him on the road, where his performances generated enough capital to send him to school in Paris and London. By age eight he was internationally famous, performing lightning calculations in England, and was described in the Annual Register as “the most singular phenomenon in the history of the human mind that perhaps ever existed.” No less than Michael Faraday and Samuel Morse admired him.

No matter where he went, Colburn met all challengers with speed and precision. He tells us in his autobiography of one set of problems he was given in New Hampshire in June 1811: “How many days and hours since the Christian Era commenced, 1811 years ago? Answered in twenty seconds: 661,015 days, 15,864,360 hours. How many seconds in eleven years? Answered in four seconds; 346,896,000.” Colburn used the same techniques described in this book to compute entirely in his head problems given to him. For example, he would factor large numbers into smaller numbers and then multiply: Colburn once multiplied 21,734 × 543 by factoring 543 into 181 × 3. He then multiplied 21,734 × 181 to arrive at 3,933,854, and finally multiplied that figure by 3, for a total of 11,801,562.

As is often the case with lightning calculators, interest in Colburn’s amazing skills diminished with time, and by the age of twenty he had returned to America and become a Methodist preacher. He died at a youthful thirty-five. In summarizing his skills as a lightning calculator, and the advantage such an ability affords, Colburn reflected, “True, the method … requires a much larger number of figures than the common Rule, but it will be remembered that pen, ink and paper cost Zerah very little when engaged in a sum.”

WHY THESE TRICKS WORK

This section is presented for teachers, students, math buffs, and anyone curious as to why our methods work. Some people may find the theory as interesting as the application. Fortunately, you need not understand why our methods work in order to understand how to apply them. All magic tricks have a rational explanation behind them, and mathemagical tricks are no different. It is here that the mathemagician reveals his deepest secrets!

In this chapter on multiplication problems, the distributive law is what allows us to break down problems into their component parts. The distributive law states that for any numbers a, b, and c:

(b + c) × a = (b × a) + (c × a)

That is, the outside term, a, is distributed, or separately applied, to each of the inside terms, b and c. For example, in our first mental multiplication problem of 42 × 7, we arrived at the answer by treating 42 as 40 + 2, then distributing the 7 as follows:

42 × 7 = (40 + 2) × 7 = (40 × 7) + (2 × 7) = 280 + 14 = 294

You may wonder why the distributive law works in the first place. To understand it intuitively, imagine having 7 bags, each containing 42 coins, 40 of which are gold and 2 of which are silver. How many coins do you have altogether? There are two ways to arrive at the answer. In the first place, by the very definition of multiplication, there are 42 × 7 coins. On the other hand, there are 40 × 7 gold coins and 2 × 7 silver coins. Hence, we have (40 × 7) + (2 × 7) coins altogether. By answering our question two ways, we have 42 × 7 = (40 × 7) + (2 × 7). Notice that the numbers 7, 40, and 2 could be replaced by any numbers (a, b, or c) and the same logic would apply. That’s why the distributive law works!

Using similar reasoning with gold, silver, and copper coins we can derive:

(b + c + d) × a = (b × a) + (c × a) + (d × a)

Hence, to do the problem 326 × 7, we break up 326 as 300 + 20 + 6, then distribute the 7, as follows: 326 × 7 = (300 + 20 + 6) × 7 = (300 × 7) + (20 × 7) + (6 × 7), which we then add up to get our answer.

As for squaring, the following algebra justifies my method. For any numbers A and d

A2 = (A + d) × (A – d) + d2

Here, A is the number being squared; d can be any number, but I choose it to be the distance from A to the nearest multiple of 10. Hence, for 772, I set d = 3 and our formula tells us that 772 = (77 + 3) × (77 − 3) + 32 = (80 × 74) + 9 = 5929. The following algebraic relationship also works to explain my squaring method:

(z + d)2 = z2 + 2zd + d2 = z(z + 2d) + d2

Hence, to square 41, we set z = 40 and d = 1 to get:

412 = (40 + 1)2 = 40 × (40 + 2) + 12 = 1681

Similarly,

(z − d)2 = z(z − 2d) + d2

To find 772 when z = 80 and d = 3,

772 = (80 − 3)2 = 80 × (80 − 6) + 32 = 80 × 74 + 9 = 5929

Chapter 3 New and Improved Products: Intermediate Multiplication

Chapter 3

New and Improved Products:
Intermediate Multiplication

Mathemagics really gets exciting when you perform in front of an audience. I experienced my first public performance in eighth grade, at the fairly advanced age of thirteen. Many mathemagicians begin even earlier. Zerah Colburn (1804–1839), for example, reportedly could do lightning calculations before he could read or write, and he was entertaining audiences by the age of six! When I was thirteen, my algebra teacher did a problem on the board for which the answer was 1082. Not content to stop there, I blurted out, “108 squared is simply 11,664!”

The teacher did the calculation on the board and arrived at the same answer. Looking a bit startled, she said, “Yes, that’s right. How did you do it?” So I told her, “I went down 8 to 100 and up 8 to 116. I then multiplied 116 × 100, which is 11,600, and just added the square of 8, to get 11,664.”

She had never seen that method before. I