Question

一个由小正方形组成的矩形网格有 $141$ 行和 $91$ 列。每个小正方形里有放两个数字的空间。Horace 和 Vera 都把从 $1$ 到 $141 \times 91 = 12{,}831$ 的数字填入网格。Horace 按行填写：他把 $1$ 到 $91$ 依次从左到右填入第 $1$ 行，把 $92$ 到 $182$ 依次从左到右填入第 $2$ 行，并如此继续到第 $141$ 行。Vera 按列填写：她把 $1$ 到 $141$ 依次从上到下填入第 $1$ 列，再把 $142$ 到 $282$ 依次从上到下填入第 $2$ 列，并如此继续到第 $91$ 列。有多少个小正方形中两人写下了相同的数字？

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?

$7$

$10$

$11$

$12$

$19$

答案：C

知识点：丢番图方程模运算区间内整数计数

难度评级：2040

解答：

在第 $r$ 行第 $c$ 列，Horace 写的是 $(r-1)\cdot 91 + c$ ，Vera 写的是 $(c-1)\cdot 141 + r$ 。令二者相等，得 $90r - 140c = -50$ ，即 $9r = 14c - 5$ 。这要求 $c \equiv 1 \pmod 9$ ，所以 $c = 1, 10, 19, \ldots, 91$ ，共有 $11$ 个值；每个都给出合法的 $r$ ，且在 $1$ 到 $141$ 之间。因此有 $11$ 个小正方形匹配。

所以正确答案是 C。

In row $r,$ column $c,$ Horace writes $(r-1)\cdot 91 + c$ and Vera writes $(c-1)\cdot 141 + r.$ Setting these equal gives $90r - 140c = -50,$ i.e. $9r = 14c - 5.$ This requires $c \equiv 1 \pmod 9,$ so $c = 1, 10, 19, \ldots, 91$ — that is $11$ values, and each yields a valid $r$ between $1$ and $141.$ So $11$ squares match.

Thus, the correct answer is C.

2025 AMC 12B 第 19 题

答案：C

难度评级：2040

解答：

其他年份的第 19 题