2000 AMC 8 第 20 题

先试着解答 2000 AMC 8 第 20 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 2000 AMC 8 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。

20.

你有九枚硬币:包括一美分硬币、五美分硬币、十美分硬币和二十五美分硬币,总价值为 $1.02\$1.02,且每种硬币至少有一枚。你必须有多少枚十美分硬币?
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02,\$1.02, with at least one coin of each type. How many dimes must you have?
11
22
33
44
55
答案:A
知识点:钱币模运算分类讨论
小提示:
用模 55 的分值确定一美分硬币的数量。
Use cents modulo 55 to find the number of pennies
大提示:
去掉一美分硬币后,再求五美分硬币、十美分硬币和二十五美分硬币的数量。
After the pennies, solve for nickels, dimes, and quarters
解答:
一美分硬币的数量必须与 102102 模 55 的余数相同,所以可能有 22 枚或 77 枚一美分硬币。如果有七枚,只剩两枚硬币给五美分硬币、十美分硬币和二十五美分硬币,不可能每种至少一枚。
所以有 22 枚一美分硬币。剩下 77 枚硬币价值 100100 分。设五美分硬币、十美分硬币和二十五美分硬币的数量分别为 nn、dd、qq,则 n+d+q=7n+d+q=7,且 5n+10d+25q=1005n+10d+25q=100。
将价值方程除以 55,再减去硬币数量方程,得到 d+4q=13d+4q=13。唯一正整数解是 q=3q=3、d=1d=1、n=3n=3。
因此必须有 11 枚十美分硬币。
所以正确答案是 A。
The number of pennies must have the same remainder as 102102 modulo 5,5, so there are either 22 or 77 pennies. Seven pennies would leave only two coins for nickels, dimes, and quarters, impossible because at least one of each type is needed.
So there are 22 pennies. The remaining 77 coins are worth 100100 cents. If n,n, d,d, and qq are the numbers of nickels, dimes, and quarters, then n+d+q=7n+d+q=7 and 5n+10d+25q=100.5n+10d+25q=100.
Dividing the value equation by 55 and subtracting the coin-count equation gives d+4q=13.d+4q=13. The only positive solution is q=3,q=3, d=1,d=1, and n=3.n=3.
Thus there must be 11 dime.
Thus, A is the correct answer.
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