Which of the following numbers is divisible by both 2 and 5?

Correct — A. A number divisible by both 2 and 5 is divisible by their LCM, 10 — so its last digit must be 0.

Test the units digit of each option:
63840 ends in 0 — even and a multiple of 5.
20792 ends in 2 — even, not a multiple of 5.

37915 ends in 5 — a multiple of 5, but odd.
37254 ends in 4 — even, not a multiple of 5.

Only 63840 clears both tests → option (a). No division is needed.

Why the others are wrong

• (b)
  20792 — 20792 ends in 2, so it passes the test for 2 and fails the test for 5, which needs a last digit of 0 or 5.
• (c)
  37915 — 37915 ends in 5, so it is a multiple of 5 but odd. An odd number is never divisible by 2, so it fails half the requirement.
• (d)
  37254 — 37254 ends in 4. It is even, and its digits sum to 21 so it is also a multiple of 3, but 5 needs a terminal 0 or 5 and this has neither.

Divisibility by 2 is decided by the last digit being even. Divisibility by 5 is decided by the last digit being 0 or 5. Only one digit satisfies both at once, and that digit is 0.

The general rule behind this: when two divisors are co-prime, passing both tests is exactly the same as being divisible by their product. Here 2 and 5 are co-prime, so the test collapses to divisibility by 10.

That is why the question is a units-digit scan, not an arithmetic exercise.

Key facts

• A number is divisible by 2 when its last digit is even, and by 5 when its last digit is 0 or 5.
• For co-prime divisors, satisfying both is identical to being divisible by their product — 2 and 5 give 10.
• Divisibility by 3 is tested on the digit sum: 37254 sums to 21, so it is a multiple of 3.

Common traps

• Checking only one of the two divisors and taking the first option that passes.
• Reading "divisible by both" as "divisible by either".
• Forgetting that 0 is the only units digit that satisfies 2 and 5 together.

No directly related past PYQ was found.