Problem 6 · 2023 AMC 8
Medium
Arithmetic & Operations
order-of-operationscasework
The digits 2, 0, 2, and 3 are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
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Answer: C — 9.
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Hint 1 of 2
The 0 is dangerous — as a base or a factor it wipes the product to 0. Where could you hide it so it does no harm?
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Hint 2 of 2
Park the 0 in an exponent (anything0 = 1), making that factor 1. Then make the other factor as big as you can from {2, 2, 3} — check both 23 and 32.
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Approach: place 0 as an exponent, maximize the rest
- The 0 is the problem child. As a base or factor it would crush the whole product to 0 — so the move is to defuse it by parking it in an exponent, where anything0 = 1 (harmless).
- That turns one factor into 1, and now you just want the other factor as big as possible using the leftover {2, 2, 3}. Compare 23 = 8 and 32 = 9 — a bigger base with a smaller exponent wins here, giving 32 = 9.
- Maximum product: 1 × 9 = 9. Worth keeping: a 0 you can't avoid is least damaging as an exponent; and when balancing base vs. exponent for small numbers, the larger base often beats the larger exponent.
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