Replace * by the smallest digit so that 723*56* is divisible by 6.

Correct — B. Both stars stand for the same digit, so call it d and split 6 into 2 × 3.

Even test: the last digit is d itself, so d must be even. That removes 1 and 3 straight away.

Digit sum: 7 + 2 + 3 + d + 5 + 6 + d = 23 + 2d
d = 2 → 27, a multiple of 3 ✓
d = 4 → 31, not a multiple of 3 ✗

The smallest even digit that also clears the 3-test is 2, giving 7232562 = 6 × 1205427 → option (b)

Why the others are wrong

• (a)
  1 — d = 1 leaves the number ending in 1, so it is odd and fails the 2-half of the test before the digit sum is worth computing.
• (c)
  3 — d = 3 is odd as well, and it misses twice over: the digit sum becomes 23 + 6 = 29, which is not a multiple of 3 either.
• (d)
  4 — d = 4 passes the even test, which is why it tempts. But the digit sum is 23 + 8 = 31, and 31 is not a multiple of 3, so the number is divisible by 2 and not by 6.

6 = 2 × 3, and 2 and 3 share no common factor, so a number is divisible by 6 exactly when it clears both tests independently.

The 2-test reads the last digit alone. The 3-test adds every digit and asks whether the total is a multiple of 3.

The twist here is that the same symbol appears twice, once inside the number and once at the end. So the unknown enters the digit sum twice — 23 + 2d, not 23 + d — and it simultaneously decides the parity.

Key facts

• A number is divisible by 6 when it is divisible by both 2 and 3.
• The test for 2 looks only at the last digit.
• The test for 3 adds all the digits and checks the sum against a multiple of 3.
• In 723*56* the unknown occupies two places, so the digit sum is 23 + 2d.

Common traps

• Counting the unknown once in the digit sum and working with 23 + d.
• Clearing the 3-test and forgetting that the last digit must also be even.
• Answering with the largest workable digit when the question asks for the smallest — 8 also works here.

No directly related past PYQ was found.