## Math Explorer's Club Tips for Mental Computations

Introduction
Lesson 2: Subtraction
Lesson 3: Multiplication
Lesson 4: Division
Lesson 5: Calendar computations
Lesson 6: Guessing a number and other tricks

# Multiplying by 11

ab × 11
ab
ab
100a + 10(a + b) + b

Suppose that a + b &le 9. In this case ab × 11 is a three-digit number whose first digit is a, second digit is a + b, and third digit is b. For instance, 36 × 11 = 396.

What if a + b &ge 10? In that case, a + 1 is the first digit (or two if a = 9), and the second digit (or third) is the last digit of a + b, and the last digit is the last is b. For instance, consider a = 9 and b = 3. Then a + 1 = 10, a + b = 12 (so 2 is the second digit) and the last digit is b = 3. Thus

93 × 11 = 1023.

Activity: ab × 11 and beyond
1. 34 × 11
2. 23 × 11
3. 39 × 11
4. 92 × 11
2. Notice that 154 × 11 = 1694. The first and last digit of 154 and 154 × 11 are the same. Moreover, notice that 1 + 5 = 6 and 5 + 4 = 9. Suppose that abc is a three digit number with a + b &le 9 and b + c &le 9. Generalize the above rule to compute abc × 11.
3. Give a rule to compute abc × 11 where either a + b &ge 10 or b + c &ge 10 (or both).
4. Do the product of a three digit number abc and 111. Can you think of a rule that makes this computation easy to do mentally?

# Products of the form ab × ac, with c = a &minus b

Let's look at some easy tricks for multiplying some numbers fast.

First consider the multiplication of two-digit numbers that start with the same digit and whose second digits sum to 10. For instance, 34 and 36. Notice that
34 × 36 = 1224.

Let's look at more examples to see if there is a pattern. Consider
45 × 45 = 2025
89 × 81 = 7209
52 × 58 = 3016

By now, we notice the following pattern: We multiply the first digit x times x + 1, and concatenate the result with the product of the last two digits.

Let's see why this trick works in the following activity. We first need to recall DECIMAL EXPANSION

Activity: ab × ac, where c = 10 &minus b.
1. Start with two two-digit numbers ab and ac, where c = 10 &minus b. Write them in their decimal expansion.
2. Show that ab × ac is the concatenation of a × (a + 1) and b × c.
3. Notice that the trick also works if we allow a to be any number (as supposed to just a digit).

# bc × de, where d = b + 1 and e = 10 &minus c

Let us consider the product of two numbers of of the form bc and de, where d = b + 1 and e = 10 &minus c. So b and d can be any numbers, but c and d are digits. For example with b = 156, c = 7 we get the pair 1563 and 1577.

There is a faster way to compute the value of the product bc × de, where d = b + 1 and e = 10 &minus c. Let's look at some examples and deduce a trick to multiply numbers of this form.

13 × 27 = 351 = 400 &minus 49 = 202 &minus 72
34 × 46 = 1564 = 1600 &minus 36 = 402 &minus 62
114 × 126 = 14364 = 14400 &minus 36 = 1202 &minus 62

Let's focus our attention to the first example: 13 + 7 = 20 = 27 &minus 7. So there is a number (7 in our case) which added to one of the factors produces the same result (20 in our case) as subtracting the same number to the other factor. Notice that this is the case for the other examples as well. In general, this number is e = 10 &minus c.

Here is the trick: bc × de = (bc + e)2 &minus e2. As an example, consider 57 × 63. Here b = 5, c = 7, d = 6, and e = 3. So (bc + e)2 &minus e2 = (57 + 3)2 &minus 32=3600 &minus 9 = 3591.

Activity: bc × de, where d = b + 1 and e = 10 &minus c
1. Show that bc + e = d × 10 = de &minus e.
2. Show that bc × de = (bc + e)2 &minus e2.
3. Compute
1. 2 × 58
2. 33 × 47
3. 1003 × 1017.

# Squaring a number

Let's get the square of 15. Notice that the previous trick can be applied to do this computation: We concatenate the results of the products 1 × 2 and 5 × 5 to get that 15 × 15 = 225. So we have a nice trick to square any number that ends with 5.

What about squaring numbers like 33? We can do this as follows: 332 = 1080 + 9 = 1089 = 30 × 36 + 32. Notice that computing 30 × 36 is as complicated as computing 3 × 36. So to compute a2 take x = a + c and y = a - c. Then,

a2 = x × y + c2.
In the example above, a = 23 and c = 3. The idea is to take c so that the product x × y is easy to compute; for instance, take c so that either x or y ends with 0.

Let's try the above technique with 1082. We can take c to be anything we want, but the computation is very easy if we take c = 8. In this case x = 116 and y = 100. So

1082 = 116 × 100 + 82
1082 = 11600 + 64 = 11664

Activity: Squaring numbers
1. Square the following numbers using the previous trick: 23, 37, 68.
2. Given a, x, y, and c as above show that a2 = x × y + c2.

# Multiplying numbers that end with 5

If one multiplies two integers x and y that end with 5 there are a couple of things we notice almost immediately. For starters, the product x × y always ends with a 5. Furthermore, the second to last digit is either a 2 or a 7; that is, x × y ends with 25 or 75.

Let's do some examples.
45 × 125 = 5625
15 × 25 = 375
65 × 95 = 6175
35 × 55 = 1925

We observe that the product ends in 25 when the sum of the numbers before 5 is even: In the first example, 4 + 12 = 16 is even and in the fourth example 3 + 5 = 8 is also even. Furthermore the product ends in 75 when the sum of the numbers before 5 is odd: In the second example, 1 + 2 = 3 is odd and in the third example 6 + 9 = 15 is also odd. This fact can be proved very easily, and you should try to prove it.

Here is a way to compute the product of two number that end with 5. Let x = a5 and y = b5. Then

1. If a + b is even, then add half of a + b to ab to get z. Then x × y = z25.
2. If a + b is odd, then add the integer part of half of a + b to ab to get z. Then x × y = z75.
Let's look at our examples: Let's look at 45 × 125. Here, a = 4 and b = 12. Since 4 + 12 = 16 is even, we apply the first rule to get z = 4 × 12 + 8 = 56. The result is 5625.

On the other hand, in 65 × 95 we take a = 6 and b = 9. So z = 6 × 9 + 7 = 61 (7 is the integer part of 6 + 9 = 15 divided by 2). So since the sum a + b is odd we apply the second rule to get 6175.

Activity: Multiplying numbers that end with 5
1. Given x = a5 and y = b5, where a and b are integers, show that x × y ends with 25 if a + b is even. Moreover, x × y ends with 75 if a + b is odd.
2. Show that the trick described above produces the correct result.
3. Using the trick described, compute
1. 25 × 15
2. 5 × 75
3. 205 × 115
Cornell University, Department of Mathematics