Find the greatest value of k for which the 6-digit number 24312k is divisible…

Given:

24312k is divisible by 6

Concept used:

If a number is even or a number whose last digit is an even number i.e. 2,4,6,8 including 0, it is always completely divisible by 2

a number is completely divisible by 3 if the sum of its digits is divisible by 3.

Calculation:

For divisible by 6 it must be divisible by both 2 and 3

24312k must be divisible by both 2 and 3

Now, possible values of k = 0, 2, 4, 6, 8

2 + 4 + 3 + 1 + 2 + 0 = 12 divisible by 3

2 + 4 + 3 + 1 + 2 + 2 = 14 not divisible by 3

2 + 4 + 3 + 1 + 2 + 4 = 16 not divisible by 3

2 + 4 + 3 + 1 + 2 + 6 = 18 divisible by 3

2 + 4 + 3 + 1 + 2 + 8 = 20 not divisible by 3

So, greatest value of k is 6

∴ Required answer is 6