If (1x2y6z) is a number divisible by 9 , then what os the least value of x+ y+z ?

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.

9514 1404 393

  0

Step-by-step explanation:

For a number to be divisible by 9, the sum of the digits must be divisible by 9. The given digits total 9, so x+y+z must be a multiple of 9. The least possible multiple of 9 is 0.

The least value of x+y+z is 0.

102060/9 = 11340

80

step-by-step explanation: