2025 AMC 12B 第 19 题

先试着解答 2025 AMC 12B 第 19 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2025 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

19.

一个由小正方形组成的矩形网格有 141141 行和 9191 列。每个小正方形里有放两个数字的空间。Horace 和 Vera 都把从 11 到 141×91=12,831141 \times 91 = 12{,}831 的数字填入网格。Horace 按行填写:他把 11 到 9191 依次从左到右填入第 11 行,把 9292 到 182182 依次从左到右填入第 22 行,并如此继续到第 141141 行。Vera 按列填写:她把 11 到 141141 依次从上到下填入第 11 列,再把 142142 到 282282 依次从上到下填入第 22 列,并如此继续到第 9191 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 141141 rows and 9191 columns. Each square has room for two numbers. Horace and Vera each fill in the grid by putting the numbers from 11 through 141×91=12,831141 \times 91 = 12{,}831 into the squares. Horace fills the grid horizontally: he puts 11 through 9191 in order from left to right into row 1,1, puts 9292 through 182182 into row 22 in order from left to right, and continues similarly through row 141.141. Vera fills the grid vertically: she puts 11 through 141141 in order from top to bottom into column 1,1, then 142142 through 282282 into column 22 in order from top to bottom, and continues similarly through column 91.91. How many squares get two copies of the same number?
77
1010
1111
1212
1919
答案:C
知识点:丢番图方程模运算区间内整数计数
小提示:
分别写出第 rr 行第 cc 列中 Horace 和 Vera 填入的数字,并将结果表示为 rr 和 cc 的公式
Write the Horace-number and the Vera-number of the square in row r,r, column cc as formulas in rr and cc
大提示:
令两式相等可得 9r=14c−59r = 14c - 5;找出哪些 cc 会使 rr 是指定范围内的整数
Setting them equal gives 9r=14c−5;9r = 14c - 5; find which cc make rr an integer in range
解答:
在第 rr 行第 cc 列,Horace 写的是 (r−1)⋅91+c(r-1)\cdot 91 + c,Vera 写的是 (c−1)⋅141+r(c-1)\cdot 141 + r。令二者相等,得 90r−140c=−5090r - 140c = -50,即 9r=14c−59r = 14c - 5。这要求 c≡1(mod9)c \equiv 1 \pmod 9,所以 c=1,10,19,…,91c = 1, 10, 19, \ldots, 91,共有 1111 个值;每个都给出合法的 rr,且在 11 到 141141 之间。因此有 1111 个小正方形匹配。
所以正确答案是 C。
In row r,r, column c,c, Horace writes (r−1)⋅91+c(r-1)\cdot 91 + c and Vera writes (c−1)⋅141+r.(c-1)\cdot 141 + r. Setting these equal gives 90r−140c=−50,90r - 140c = -50, i.e. 9r=14c−5.9r = 14c - 5. This requires c≡1(mod9),c \equiv 1 \pmod 9, so c=1,10,19,…,91c = 1, 10, 19, \ldots, 91 — that is 1111 values, and each yields a valid rr between 11 and 141.141. So 1111 squares match.
Thus, the correct answer is C.
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