Problem 7 · 2018 AMC 8
Medium
Number Theory
divisibilitydigit-sum
The 5-digit number 2018U is divisible by 9. What is the remainder when this number is divided by 8?
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Answer: B — Remainder 3.
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Hint 1 of 2
The mystery digit isn't really free. The divisible-by-9 rule keys off the digit sum, and a single missing digit can only nudge that sum into one nearby multiple of 9 — so U is pinned down before you do anything else.
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Hint 2 of 2
The technique: use a divisibility rule to solve for the unknown digit first, then treat the now-known number as a plain division problem.
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Approach: use the divisibility-by-9 rule
- Add the known digits: 2 + 0 + 1 + 8 = 11. The full sum 11 + U must be a multiple of 9, and since U is a single digit (0–9) the only reachable multiple is 18, forcing U = 7.
- Now the number is 20187. Divide by 8: 2523 × 8 = 20184, leaving a remainder of 3.
- You'll see it again: "unknown digit makes it divisible by 9 (or 3)" problems always start with the digit-sum rule — it converts an unknown digit into a forced one.
Another way — remainder mod 8 uses only the last three digits:
- A divisibility shortcut for 8: a number's remainder when divided by 8 depends only on its last three digits, because 1000 is a multiple of 8.
- So instead of dividing 20187, just look at 187: 187 = 23×8 + 3, giving remainder 3 — the same answer with much smaller arithmetic.
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