15 August 2011 - 1 Multiplying Numbers When the First Digits are the Same & the Last Sum to Ten

Continuing the theme of multiplying various numbers, we come to some which satisfy a particular condition as explained in the examples below.

Example 1

We'll multiply 26 by 24. The method we'll use requires the first digits to be the same, here 2, and the last digit from each number sums to ten, here 6 + 4 = 10.

1. Take the first digit	2
2. Add one	2 + 1 = 3
3. Multiply the numbers from step 1 and 2.	2 x 3 = 6
4. Multiply the last 2 digits of the original numbers	6 x 4 = 24
5. Write the answer to step 4 after step 3.	624

Hence 26 x 24 = 624. The method also works for larger numbers.

Example 2

We'll multiply 113 by 117. We'll check the numbers satisfy the conditions for the method. The first digits are the same, both 11, and the last two digits sum to ten: 3 + 7 = 10.

1. Take the first digit	11
2. Add one	11 + 1 = 12
3. Multiply the numbers from step 1 and 2.	11 x 12 = 132
4. Multiply the last 2 digits of the original numbers	3x7=21
5. Write the answer to step 4 after step 3.	13221

Hence 113 x 117 = 13221.

Proof

Again, let's check that the examples above weren't just flukes. Let's multiply two numbers a and b which both start with c which could be a single digit or several digits, and end in d and (10 - d) respectively. Hence

a = 10c + d and b = 10c + (10-d).

Multiplying a and b, we get

ab = (10c + d)(10c + (10-d)) 
= 100c²+ 100c -10cd + 10cd + 10d - d²

= 100c²+ 100c + 10d - d².

We'll now check to see if the method gives the same result.

1. Take the first digit	c
2. Add one	c + 1
3. Multiply the numbers from step 1 and 2.	c(c + 1) = c2 + c
4. Multiply the last 2 digits of the original numbers	d(d-10) = d2 - 10d
5. Write the answer to step 4 after step 3. This is the equivalent to multiplying the answer to step 3 by 10 and then adding the answer to step 4.	10(c2 + c)+d2 - 10d = 10c2 + 10c + d2 - 10d

Hence the method gives 10c² + 10c + d² - 10d which is ab so it works for numbers of the prescribed form.

Source: http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp