Which of the following can be the value of 'k' so that the number 217924k is divisible by 6?

Question

UPSC Quizroom · Accepted Answer

A number is divisible by 6 if:

It is divisible by 2

It is divisible by 3

Number: 217924k

Divisibility by 2:
Last digit must be even → k = 0, 2, 4, 6, 8

Divisibility by 3:
Sum of digits = 2 + 1 + 7 + 9 + 2 + 4 + k = 25 + k

For divisibility by 3:
25 + k ≡ 0 (mod 3)
25 mod 3 = 1
So: 1 + k ≡ 0 (mod 3)
k ≡ 2 (mod 3)

From even digits {0,2,4,6,8}, only k = 2 and k = 8 satisfy k ≡ 2 (mod 3).

Final Answer: k = 2 or 8

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