If 321y72 is a multiple of 6, where y is a digit, what is the least value of y?

Correct — A. A multiple of 6 must pass both tests: divisible by 2 and by 3.

By 2: the last digit of 321y72 is 2, so this holds whatever y is.
By 3: digit sum = 3 + 2 + 1 + y + 7 + 2 = 15 + y

15 is already a multiple of 3, so 15 + y is one when y = 0, 3, 6 or 9.
The least of those is y = 0 → option (a)

Check: 321072 ÷ 6 = 53,512 exactly.

Why the others are wrong

• (b)
  7 — 7 makes the digit sum 15 + 7 = 22, which is not a multiple of 3. The number stays even but fails the second test, so it is not a multiple of 6.
• (c)
  2 — 2 makes the digit sum 17, again not a multiple of 3. 321272 is even and still not divisible by 6.
• (d)
  3 — 3 does work — the digit sum becomes 18 — but the stem asks for the least value of y, and 0 is smaller and equally valid.

Six is not prime, so it has no rule of its own. Split it into coprime factors 2 and 3 and apply both rules; the same trick turns 12 into 3 and 4, and 15 into 3 and 5.

Divisibility by 2 depends only on the last digit, which is fixed at 2 here, so that test is already passed and y is free of it.

Divisibility by 3 depends only on the digit sum. The known digits add to 15, itself a multiple of 3, so y must be a multiple of 3 as well — and 0 is a multiple of 3.

Key facts

• A number is divisible by 6 exactly when it is divisible by both 2 and 3.
• Divisibility by 3 depends only on the digit sum, and 3 + 2 + 1 + 7 + 2 = 15.
• The digits that make 15 + y a multiple of 3 are 0, 3, 6 and 9, of which 0 is least.
• Zero is a multiple of 3, since 3 × 0 = 0.

Common traps

• Rejecting y = 0 on the belief that a digit cannot be zero, when y is not the leading digit
• Choosing 3, the smallest non-zero digit that works
• Checking only the digit-sum rule and never confirming the number is even

No directly related past PYQ was found.