Lesson 1: Addition
Lesson 2: Subtraction
Lesson 3: Multiplication
Lesson 4: Division
Lesson 5: Calendar computations
Lesson 6: Guessing a number and other tricks
Multiplying by 11Let us start with a two-digit number ab. Notice that
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.
Activity: ab × 11 and beyond
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
Let's look at more examples to see if there is a pattern. Consider
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.
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.
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
Squaring a numberLet'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,
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 = 11600 + 64 = 11664
Activity: Squaring numbers
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.
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
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