2021 AMC 10B Fall 第 7 题

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7.

如果分数 ab\frac{a}{b} 不一定为最简形式,且 aabb 为正整数、两者之和为 1515,则称它为“特殊”分数。有多少个不同的整数可以写成两个不一定不同的特殊分数之和?

Call a fraction ab,\frac{a}{b}, not necessarily in simplest form, special if aa and bb are positive integers whose sum is 15.15. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

 9\ 9

 10\ 10

 11\ 11

 12\ 12

 13\ 13

答案:C
知识点:分数整除性系统列举
难度评级:1660
解答:

分母为 bb 的特殊分数等于 15bb=15b1\frac{15-b}{b}=\frac{15}{b}-1,其中 1b141\le b\le14。所求整数可写成 15x+15y2\frac{15}{x}+\frac{15}{y}-2

逐一检查可能分母的小数部分,得到不同的整数和为 xyx\le y1,2,3,4,6,7,8,13,16,18,281,2,3,4,6,7,8,13,16,18,28(1,1),(1,3),(1,5),(2,2),(2,6),(2,10),(3,3),(3,5),(4,12),(5,5),(6,6),(6,10),(10,10). \begin{gathered} (1,1),(1,3),(1,5),(2,2),(2,6),\\ (2,10),(3,3),(3,5),(4,12),\\ (5,5),(6,6),(6,10),(10,10). \end{gathered}

这样的整数共有 1111 个。

所以答案是 C

A special fraction with denominator bb equals 15bb=15b1,\frac{15-b}{b}=\frac{15}{b}-1, where 1b14.1\le b\le14. We need integer values of 15x+15y2.\frac{15}{x}+\frac{15}{y}-2.

Taking xy,x\le y, a check of the fourteen possible denominators gives the following pairs that produce integers: (1,1),(1,3),(1,5),(2,2),(2,6),(2,10),(3,3),(3,5),(4,12),(5,5),(6,6),(6,10),(10,10). \begin{gathered} (1,1),(1,3),(1,5),(2,2),(2,6),\\ (2,10),(3,3),(3,5),(4,12),\\ (5,5),(6,6),(6,10),(10,10). \end{gathered} Their distinct sums are 1,2,3,4,6,7,8,13,16,18,28.1,2,3,4,6,7,8,13,16,18,28.

There are 1111 such integers.

Thus, the answer is C .

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