16 . Playing with Numbers

Chapter 16 . Playing with Numbers

16. Playing with Numbers

Exercise 16.1

Find the values of the letters in each of the following and give reasons for the steps involved.

Solution: We have ,

This has two letters A and B whose values are to be found .

If A = 7 , then the sum is 7 + 5 = 12 .

So, B = 3 + 2 + Carried on = 3 + 2 + 1 = 6

Therefore ,

So, the answer is A = 7 and B = 6 .

Solution: We have ,

This has two letters A , B and C whose values are to be found .

If A = 5 , then the sum is A + 8 = 5 + 8 = 13 .

The value of B and C = 4 + 9 + Carried on = 4 + 9 + 1 = 14

Therefore ,

So, the answer is A = 5 , B = 4 and C = 1 .

Solution: We have ,

This has one letter whose values are to be found .

Since the ones digit of A × A is A , it must be that A = 1 or A = 6 .

If A = 1 , then A × A = 1 × 1 = 1 , then .

If A = 6 , then A × A = 6 × 6 = 36 . So, .

Therefore ,

So, the value of A = 6

Solution: We have ,

This has two letters A and B whose values are to be found .

If B = 5 , then the sum is 7 + B = 7 + 5 = 12 .

If A = 2 , then the sum is 3 + A + carried on = 3 + 2 + 1 = 6 .

Therefore ,

So, the answer is A = 2 and B = 5 .

Solution: We have ,

This has two letters A and B whose values are to be found .

Since the ones digit of 3 × B is B , it must be that B = 0 .

If B = 0 , then 3 × B = 3 × 0 = 0 .

If A = 5 , then 3 × A = 3 × 5 = 15 .

Therefore ,

So, the answer is A = 5 , B = 0 and C = 1 .

Solution: We have ,

This has three letters A , B and C whose values are to be found .

Since the ones digit of 5 × B is B , it must be that B = 0 .

If B = 0 , then 5 × 0 = 5 × 0 = 0 .

If A = 5 , the 5 × A = 5 × 5 = 25 .

Therefore ,

So, the answer is A = 5 , B = 0 and C = 2 .

Solution: We have ,

This has one letter whose values are to be found .

Since the ones digit of 6 × B is B , it must be that B = 4 .

If B = 4 , then 6 × B = 6 × 4 = 24 .

If A = 7 , then 6 × A = 6 × 7 = 42 .So , .

Therefore ,

So, the answer is A = 7 and B = 4 .

Solution: We have ,

This has two letters A and B whose values are to be found .

If B = 9 , then the sum is 1 + B = 1 + 9 = 10 .

If A = 7 , then the sum is 1 + A + carried on = 1 + 7 + 1 = 9 .

Therefore ,

So, the answer is A = 7 and B = 4 .

Solution: We have ,

This has two letters A and B whose values are to be found .

If B = 7 , then the sum is 1 + B = 1 + 7 = 8 .

If A = 4 , then the sum is A + B = 4 + 7 = 11 .

So, B = 2 + A + carried on = 2 + 4 + 1 = 7 .

Therefore ,

So, the answer is A = 4 and B = 7 .

10.

Solution: We have ,

This has two letters A and B whose values are to be found .

If A = 8 and B = 1 , then the sum is A + B = 8 + 1 = 9 .

So , A + 2 = 8 + 2 = 10 .

Again, 1 + 6 + carried on = 1 + 6 + 1 = 8 .

Therefore ,

So, the answer is A = 8 and B = 1 .

Exercise 16.2

1. If

is a multiple of 9, where

is a digit, what is the value of

Solution: Since , is a multiple of 9 .

The sum of the digits of is .

If , then the digit is is multiple of 9 .

So, the value of .

2. If

is a multiple of 9, where

is a digit, what is the value of

? You will find that there are two answers for the last problem. Why is this so?

Solution: Since , is a multiple of 9 .

The sum of the digits of is .

If , then the digit is is divisible by 9 .

So, the value of or .

3. If

is a multiple of 3, where

is a digit, what is the value of ? (Since

is a multiple of 3, its sum of digits

is a multiple of 3; so

is one of these numbers: 0, 3, 6, 9, 12, 15, 18, ... . But since x is a digit, it can only be that

or 9 or 12 or 15. Therefore,

= 0 or 3 or 6 or 9. Thus,

can have any of four different values.)

Solution: Since , is a multiple of 3 .

The sum of the digits of is .

If , then the digit is is divisible by 3 .

So, the value of or .

4. If

is a multiple of 3, where

is a digit, what might be the values of

Solution: Since , is a multiple of 3 .

The sum of the digits of is .

If , then the digit is is divisible by 3 .

So, the value of or .