Question

$27$ 块砖（长方体）的尺寸都是 $a \times b \times c$ ，其中 $a$ ， $b$ 和 $c$ 是两两互质的正整数。这些砖被排成一个 $3 \times 3 \times 3$ 的长方体块，如下图左侧所示。再加入第 $28$ 块同样尺寸的砖，并把这些砖重新排成一个 $2 \times 2 \times 7$ 的长方体块，如右侧所示。新的长方体比旧的高 $1$ 个单位、宽 $1$ 个单位、深 $1$ 个单位。求 $a + b + c$ 。

Each of $27$ bricks (right rectangular prisms) has dimensions $a \times b \times c,$ where $a,$ $b,$ and $c$ are pairwise relatively prime positive integers. These bricks are arranged to form a $3 \times 3 \times 3$ block, as shown on the left below. A $28$ th brick with the same dimensions is introduced, and these bricks are reconfigured into a $2 \times 2 \times 7$ block, shown on the right. The new block is $1$ unit taller, $1$ unit wider, and $1$ unit deeper than the old one. What is $a + b + c?$

$88$

$89$

$90$

$91$

$92$

答案：E

知识点：丢番图方程长方体分类讨论

难度评级：2470

小提示：

旧块的边长为 $3a, 3b, 3c$ ；新块的边长为 $2u, 2v, 7w$ ，其中 $u, v, w$ 是 $a, b, c$ 的某种排列。

The old block has side lengths $3a, 3b, 3c;$ the new block has side lengths $2u, 2v, 7w$ for some assignment of $a, b, c$ to $u, v, w$

大提示：

重新标号，使新块的边长为 $7a,2b,2c$ ；其中任何一条都不可能等于字母相同的那个 $3a+1,3b+1,3c+1$ 。

Relabel so the new side lengths are $7a,2b,2c;$ no one of these can equal $3a+1,3b+1,3c+1$ with the same letter

解答：

重新标记砖块的尺寸，使新块的边长为 $7a,2b,2c$ 。这三条边必定就是旧块的三条边长 $3a,3b,3c$ 各增加 $1$ 之后的结果。新边不可能与字母相同的旧边对应： $7a=3a+1$ 没有正整数解，而 $2b=3b+1$ 和 $2c=3c+1$ 会给出负的边长。因此这个对应必定是两个三轮换之一。在其中一种排法下， $\begin{aligned} 7a&=3c+1,\\ 2b&=3a+1,\\ 2c&=3b+1 \end{aligned}\text{。}$ 后两个方程给出 $b=\frac{3a+1}{2}$ 和 $c=\frac{9a+5}{4}$ ，代入第一个方程得 $28a=27a+19$ ，所以 $(a,b,c)=(19,29,44)$ 。另一个三轮换只是把 $b$ 和 $c$ 互换。这三个长度两两互质，且 $a+b+c=19+29+44=92$ 。因此正确答案是 E。

Relabel the brick dimensions so the new block has sides $7a,2b,2c.$ These must be the three old side lengths $3a,3b,3c,$ each increased by $1.$ A new side cannot match the old side with the same letter: $7a=3a+1$ has no positive integer solution, while $2b=3b+1$ and $2c=3c+1$ would give negative lengths. Therefore the matching must be one of the two three-cycles. In one orientation, $\begin{aligned} 7a&=3c+1,\\ 2b&=3a+1,\\ 2c&=3b+1. \end{aligned}$ The last two equations give $b=\frac{3a+1}{2}$ and $c=\frac{9a+5}{4}.$ Substituting into the first gives $28a=27a+19,$ so $(a,b,c)=(19,29,44).$ The other cycle merely exchanges $b$ and $c.$ These lengths are pairwise relatively prime, and $a+b+c=19+29+44=92.$ Thus, E is the correct answer.

2024 AMC 10B 第 25 题

答案：E

难度评级：2470

小提示：

大提示：

解答：

其他年份的第 25 题