πŸ‡ΊπŸ‡Έ AMC 8 ⇄ switch contest
2000 AMC 8

Problem 9

Problem 9 · 2000 AMC 8 Medium
Number Theory powerscareful-counting
Figure for AMC 8 2000 Problem 9
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Answer: D — 6.
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Hint 1 of 2
The clue 'three-digit power' is doing the heavy lifting β€” there are only a handful of those, so list them instead of computing. The two answers must agree at the square where they cross.
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Hint 2 of 2
Powers of 5 grow fast: only 125 and 625 have three digits. Powers of 2: only 128, 256, 512. The crossing cell must be a digit that appears in BOTH an answer of one list and an answer of the other.
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Approach: list the few candidates, then match at the shared cell
  1. Three-digit powers of 5 are just 125 and 625 β€” both have 2 as their *middle* digit. So the down-answer's middle cell (which is the across-answer's *first* cell) is a 2.
  2. Three-digit powers of 2 are 128, 256, 512. The only one starting with 2 is 256, so the across answer is 256.
  3. The outlined square is the last cell of that across answer: its units digit, 6.
  4. The lesson: 'how many such numbers exist' is often tiny β€” exponential values thin out fast, so enumerate them rather than search blindly, then let the overlap pin down the rest.
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