AMC10 2025 B

AMC10 2025 B · Q23

AMC10 2025 B · Q23. It mainly tests GCD & LCM, Diophantine equations (integer solutions).

A rectangular grid of squares has $141$ rows and $91$ columns. Each square has room for two numbers. Horace and Vera each fill in the grid by putting the numbers from $1$ through $141 \times 91 = 12{,}831$ into the squares. Horace fills the grid horizontally: he puts $1$ through $91$ in order from left to right into row $1$, puts $92$ through $182$ into row $2$ in order from left to right, and continues similarly through row $141$. Vera fills the grid vertically: she puts $1$ through $141$ in order from top to bottom into column $1$, then $142$ through $282$ into column $2$ in order from top to bottom, and continues similarly through column $91$. How many squares get two copies of the same number?

一个矩形方格网格有 $141$ 行和 $91$ 列。每个方格可容纳两个数字。Horace 和 Vera 各自填充网格，将 $1$ 到 $141 \times 91 = 12{,}831$ 的数字放入方格。Horace 横向填充：第 $1$ 行从左到右放 $1$ 到 $91$，第 $2$ 行放 $92$ 到 $182$，依此类推到第 $141$ 行。Vera 纵向填充：第 $1$ 列从上到下放 $1$ 到 $141$，第 $2$ 列放 $142$ 到 $282$，依此类推到第 $91$ 列。有多少个方格得到两个相同的数字？

(A) 7 7

(B) 10 10

(D) 12 12

(E) 19 19

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let's say $n$ is one of the numbers that got written twice in the same box. Suppose it is at column $x$ and row $y$. We will write an expression for $n$ in terms of $x$ and $y$ in two ways: from Horace's perspective and Vera's perspective. From Horace's perspective, $n = (y-1)(141) + x$. This is because there are $y-1$ full rows before $n$, and we then need $x$ more numbers to reach $n$. Similarly, Vera will say $n = (x-1)(91) + y$. We now have the Diophantine equation \[(y-1)(141) + x = (x-1)(91)+y\] \[141y-141+x = 91x-91+y\] \[140y=90x+50\] \[14y = 9x + 5\] One such solution is, of course, $x=y=1$. We find further solutions by adding $14$ to $x$ and $9$ to $y$. For example, the second solution is $(15,10)$. This continues until $(141,91)$ is reached. There are $\boxed{11}$ ordered pairs in this list.

设 $n$ 是同一个方格被写两次的数字。假设它在第 $x$ 列第 $y$ 行。我们用 Horace 和 Vera 的视角写出 $n$ 关于 $x$ 和 $y$ 的两种表达式。从 Horace 视角，$n = (y-1)(141) + x$。因为 $n$ 前有 $y-1$ 整行，再加 $x$ 个数字。类似地，Vera 视角 $n = (x-1)(91) + y$。现在有不定方程 \[(y-1)(141) + x = (x-1)(91)+y\] \[141y-141+x = 91x-91+y\] \[140y=90x+50\] \[14y = 9x + 5\] 一个解当然是 $x=y=1$。进一步解通过 $x$ 加 $14$、$y$ 加 $9$ 得到。例如第二个解是 $(15,10)$。继续直到达到 $(141,91)$。有序对共有 $\boxed{11}$ 个。

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