2004 AIME II 第 3 题

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3.

一个实心长方体由 NN 个全等的 11 厘米立方体面贴面粘成。当从能看见它三个面的方向观察时,恰有 23123111 厘米立方体看不见。求 NN 的最小可能值。

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.

答案:384
知识点:长方体质因数分解最优化
难度评级:2110
解答:

设长方体尺寸为 p×q×rp \times q \times r。一个小立方体看不见,当且仅当它不接触三个可见面中的任何一个, 所以看不见的小立方体形成 (p1)×(q1)×(r1)(p-1) \times (q-1) \times (r-1) 的块。因此 (p1)(q1)(r1)=231(p-1)(q-1)(r-1) = 231 =3711= 3 \cdot 7 \cdot 11

231231 写成三个正整数乘积的方式为 37113 \cdot 7 \cdot 1113771 \cdot 3 \cdot 7717331 \cdot 7 \cdot 33111211 \cdot 11 \cdot 21112311 \cdot 1 \cdot 231。对应长方体为 4×8×124 \times 8 \times 122×4×782 \times 4 \times 782×8×342 \times 8 \times 342×12×222 \times 12 \times 222×2×2322 \times 2 \times 232,体积分别为 384384624624544544528528928928

最小值为 N=384N = 384

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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