Find the greatest value of ‘k’ in a 6-digit number 6745k2 such that the number is divisible by 3.

Correct — D. A number is divisible by 3 when its digit sum is divisible by 3.

Digit sum of 6745k2 = 6 + 7 + 4 + 5 + k + 2
= 24 + k

24 is already a multiple of 3, so 3 must divide k on its own.
Single digits that work: 0, 3, 6, 9.

The stem asks for the greatest, so k = 9 → option (d).
Check: 24 + 9 = 33 = 3 × 11.

Why the others are wrong

• (a)
  7 — k = 7 gives a digit sum of 24 + 7 = 31, and 31 is not a multiple of 3 (30 is). So 674572 fails the test outright.
• (b)
  8 — k = 8 gives 32, one past 31 and still not a multiple of 3. Only digits that are themselves multiples of 3 keep 24 + k divisible by 3.
• (c)
  6 — k = 6 does work, since 24 + 6 = 30. But the stem asks for the greatest such digit, and 9 works too — this is the trap the option set is built around.

Divisibility by 3 depends only on the sum of the digits — not on their order, not on the last digit. That is why 6745k2 collapses to 24 + k and the rest of the number stops mattering.

The same style of test covers 9 (digit sum divisible by 9), 2, 5 and 10 (last digit) and 11 (alternating sum).

When a missing-digit item asks for the greatest or the least value, solve the condition first — here k must be a multiple of 3 — and only then take the end of the list the stem wants.

Key facts

• 6745k2 is divisible by 3 exactly when 6 + 7 + 4 + 5 + k + 2 = 24 + k is divisible by 3.
• 24 is a multiple of 3, so the valid digits are k = 0, 3, 6, 9 and the greatest is 9.
• Digit-sum tests decide divisibility by 3 and 9 only. For 2, 4, 5, 8 and 10 the trailing digits decide.

Common traps

• Stopping at the first digit that works: 6 satisfies the rule, but 9 is larger and also satisfies it
• Applying the digit-sum test to divisibility by 4 or 8, where only the trailing digits matter

No directly related past PYQ was found.