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.
