2004 AIME II Problem 3

Attempt Problem 3 of the 2004 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AIME II solutions, or check the answer key.

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3.

A solid rectangular block is formed by gluing together NN congruent 11-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly 231231 of the 11-cm cubes cannot be seen. Find the smallest possible value of N.N.

Answer: 384
Concepts:rectangular prismprime factorizationoptimization
Difficulty rating: 2110
Solution:

Let the block measure p×q×r.p \times q \times r. A cube is hidden exactly when it touches none of the three visible faces, so the hidden cubes form a (p1)×(q1)×(r1)(p-1) \times (q-1) \times (r-1) block, giving (p1)(q1)(r1)=231(p-1)(q-1)(r-1) = 231 =3711.= 3 \cdot 7 \cdot 11.

The ways to write 231231 as a product of three positive integers are 3711,3 \cdot 7 \cdot 11, 1377,1 \cdot 3 \cdot 77, 1733,1 \cdot 7 \cdot 33, 11121,1 \cdot 11 \cdot 21, and 11231,1 \cdot 1 \cdot 231, giving blocks 4×8×12,4 \times 8 \times 12, 2×4×78,2 \times 4 \times 78, 2×8×34,2 \times 8 \times 34, 2×12×22,2 \times 12 \times 22, and 2×2×232,2 \times 2 \times 232, with volumes 384,384, 624,624, 544,544, 528,528, and 928.928.

The smallest is N=384.N = 384.

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