🇺🇸 AMC 8 ⇄ switch contest
2026 AMC 8

Problem 16

Problem 16 · 2026 AMC 8 Hard
Number Theory divisibility-rule

Consider all positive four-digit integers whose digits are all even. What fraction of these integers are divisible by 4?

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Answer: D — 3/5.
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Hint 1 of 2
Don't count the thousands of numbers. Divisibility by 4 cares only about the last two digits — so the first two even digits are free and just cancel out of the fraction.
Still stuck? Show hint 2 →
Hint 2 of 2
Here's the lucky break: the tens digit is even, so 10×(tens) is a multiple of 20, hence already a multiple of 4. That leaves the units digit as the only thing that decides divisibility by 4.
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Approach: divisibility by 4 collapses onto the units digit alone
  1. Divisibility by 4 depends only on the last two digits, so the thousands and hundreds digits don't matter — whatever fraction works for the last two digits is the answer for the whole pile.
  2. Split the last two digits as 10·(tens) + units. The tens digit is even, so 10·(tens) is a multiple of 20, which is already a multiple of 4. That means the number is divisible by 4 exactly when the units digit alone is — i.e. units ∈ {0, 4, 8}.
  3. Of the 5 allowed even units {0, 2, 4, 6, 8}, three (0, 4, 8) work, and this holds no matter what the other digits are. So the fraction is 3/5 → 3/5.
  4. Why this transfers: a divisibility rule lets you ignore most digits — isolate just the digits the rule depends on, and check whether the ‘free’ part is automatically handled so the count reduces to one digit.
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