Question

1X2Y6Z is a number divisible by 9 then the least value of X+Y+Z is
(b) 1
(a) o​

Answer 1

Given:

Number 1X2Y6Z is divisible by 9.

To find:

The least value of  X+Y+Z = ?

Solution:

First of all, let us learn the divisibility rule of 9.

A given number will be divisible by 9 if and only if the sum of digits of the given number is also divisible by 9.

For example:

Let us consider a number 923463.

The sum of digits = 9 + 2 + 3 + 4 + 6 + 3 = 27

27 is divisible by 9 so the number 923463 will also be divisible by 9.

Let us consider another example 8972.

Sum of digits = 26

26 is not divisible by 9 so the number 8972 will also be not divisible by 9.

Now, we are given the number 1X2Y6Z.

Sum of digits = 1 + X + 2 + Y + 6 + Z = 9 + X+Y+Z

The number 1X2Y6Z will be divisible by if 9+X+Y+Z is divisible by 9.

The least value of X+Y+Z can be zero, so the sum of digits = 9 + 0 = 9

which will make the number 1X2Y6Z divisible by 9.

So, the answer is:

Least value of X+Y+Z = 0 so that the number 1X2Y6Z is divisible by 9.

Answer 2

Answer:

0 is the correct answer given upside