Chapter 13: Playing with Numbers

Solution: False

Explanation:

From the given question, the value of B should be either 3 or 8.

If you take the value of B is 3, it should be equal to 19.

Hence, the value of B is 8.

We know that, the value of A should be between 0 and 9.

Hence, A x 4 = 19- 3 = 16

A = 16/4

A = 4

Therefore A= 4 and B= 8

Hence, A+B = 4+8 = 12

Solution: True

Explanation:

From the given number, B+C is either 2 or the two-digit number that gives the unit digit as 2.

It is given that B ≠ 0, C ≠ 0,

If B or C = 5 or 7, then A should be 3, then it becomes A+B+C = 14

Or else, if B= C = 6, and A = 2, we will get A+B+C = 14

Solution: False

Explanation:

If 213 x 27 is divisible by 9, then the sum of the digits of a number is a multiple of 9.

2+1+3+x+2+7 = 15+x

Then 15x must be any multiple of 9, such as 9, 18, 27 …

Now assume that,

15+x = 18

X = 18 -15

X = 3

Solution: False

Explanation:

Given that, when N is divided by 5, it leaves the remainder 5. (i.e) N = 5n+3 where n= 0, 1, 2, 3, …

Similarly, when N is divided by 2, it leaves the remainder 0. So N is an even Number. (Using divisibility test rule of 2).

But in N = 5n+3, the second term is odd.

So, 5n is an odd number.

When you substitute n = 1, 3, 5 … in 5n+3, we will get 8, 18, 28 …

Now, if we divide N by 10, it should be written as

N = 10 n+8

So, when N is divided by 10, it always leaves the remainder 8.

Solve the following:

Solution: a = 3

Explanation:

By using the divisibility test of 8, if a number is divisible by 8, then the last three digits of a number should be divisible by 8.

In the given question, the last three digit is 6a2.

The value of “a” varies from 0 to 9.

If a = 0, then 6a2 = 602 is not divisible by 8.

If a = 1, then 612 = 612 is not divisible by 8.

If a = 2, then 6a2 = 622 is not divisible by 8.

If a = 3, then 6a2 = 632 is divisible by 8.

Hence, the least value of “a” is 3.