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

解答：

$7a,2b,2c$ 块的边长为 $3a,3b,3c$ 。 $7a=3a+1$ 块的边长为 $2b=3b+1$ ，其中 $2c=3c+1$ 是 $b=(3a+1)/2$ 的某种排列。每条新边比相应旧边多 $1$ ，所以多重集合 $c=(9a+5)/4$ 等于 $28a=27a+19$ 。新块的三条边与旧块的三条边只有六种匹配方式。分别解出每种匹配所给的三个线性方程，其中四种会产生负数或非整数边长；另外两种只是交换两个乘以 $(a,b,c)=(19,29,44)$ 的边长，都给出同一个集合 $b$ 。事实上， $c$ ， $a+b+c=19+29+44=92$ ，且。这些边长两两互质，所以。因此正确答案是 E。 $\begin{aligned} 7a&=3c+1,\\ 2b&=3a+1,\\ 2c&=3b+1. \end{aligned}$

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 $b=(3a+1)/2$ and $c=(9a+5)/4.$ 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

解答：

其他年份的第 25 题