AMC12 2025 B

AMC12 2025 B · Q19

AMC12 2025 B · Q19. 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$182$，依此类推至第$141$行。Vera纵向填充：第$1$列从上到下放$1$到$141$，第$2$列放$142$到$282$，依此类推至第$91$列。有多少方格得到两个相同数字？

(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$n$是同一格写两次的数字。假设在第$x$列第$y$行。从Horace视角，$n = (y-1)(141) + x$，因为前$y-1$行满，加上$x$个。从Vera视角，$n = (x-1)(91)+y$。得二元一次不定方程 \[(y-1)(141) + x = (x-1)(91)+y\] \[141y-141+x = 91x-91+y\] \[140y=90x+50\] \[14y = 9x + 5\] 一解为$x=y=1$。进一步解为$x$加$14$，$y$加$9$，如第二解$(15,10)$。直至达到$(141,91)$。共有$\boxed{11}$个有序对。

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