AMC10 2024 B

AMC10 2024 B · Q25

AMC10 2024 B · Q25. It mainly tests Systems of equations, 3D geometry (volume).

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 $a + b + c$?

$27$块砖（直角长方体），尺寸为$a \times b \times c$，其中$a$、$b$、$c$两两互质的正整数。这些砖排列成如下左图所示的$3 \times 3 \times 3$块。引入第$28$块相同尺寸的砖，并重新配置成右图所示的$2 \times 2 \times 7$块。新块比旧块高$1$单位、宽$1$单位、深$1$单位。求$a + b + c$？

(A) 88 88

(B) 89 89

(D) 91 91

(E) 92 \qquad 92 \qquad

Answer

Correct choice: (E)

正确答案：（E）

Solution

The $3$x$3$x$3$ block has side lengths of $3a, 3b, 3c$. The $2$x$2$x$7$ block has side lengths of $2b, 2c, 7a$. We can create the following system of equations, knowing that the new block has $1$ unit taller, deeper, and wider than the original: \[3a+1 = 2b\] \[3b+1=2c\] \[3c+1=7a\] Adding all the equations together, we get $b+c+3 = 4a$. Adding $a-3$ to both sides, we get $a+b+c = 5a-3$. The question states that $a,b,c$ are all relatively prime positive integers. Therefore, our answer must be congruent to $2 \pmod{5}$. The only answer choice satisfying this is $\boxed{E(92)}$.

$3\times3\times3$块的边长为$3a, 3b, 3c$。$2\times2\times7$块的边长为$2b, 2c, 7a$。由新块比原块高、深、宽各$1$单位，得方程组： \[3a+1 = 2b\] \[3b+1=2c\] \[3c+1=7a\] 三式相加，得$b+c+3 = 4a$。两边加$a-3$，得$a+b+c = 5a-3$。题目称$a,b,c$两两互质正整数。因此，答案必模$5$同余$2$。唯一满足的选项为$\boxed{E(92)}$。

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