Question

若分数 $\dfrac{a}{b}$ 的正整数分子 $a$ 与分母 $b$ 之和为 $15$ 则称该分数为特殊分数，分数不必是最简形式。有多少个不同的整数可以表示为两个特殊分数之和？这两个分数可以相同。

Call a fraction $\dfrac{a}{b},$ not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15.$ How many distinct integers can be written as the sum of two, not necessarily different, special fractions?

$9$

$10$

$11$

$12$

$13$

答案：C

知识点：分数分类讨论

难度评级：1400

解答：

约分后，特殊分数中的整数有 $a/(15-a)$ $1\le a\le14$ $2,4,14$ 半整数有 $1/2$ $\tfrac12,\tfrac32,\tfrac{13}{2}$ $\tfrac14,\tfrac{11}{4}$ 小数部分为四分之一或四分之三的有 $1/4,3/4$ 和此外还有其他分数。

两个特殊分数之和为整数，当且仅当它们的小数部分之和为整数。 $1/14,2/13,4/11,2/3,7/8,1/7$ ，

整数对给出 $4, 6, 8, 16, 18, 28$ 。半整数对 $\left(\tfrac12, \tfrac32, \tfrac{13}{2}\right)$ 给出 $1, 2, 3, 7, 8, 13$ 。四分之一型的一对 $\tfrac14 + \tfrac{11}{4}$ 给出 $3$ 。

去重后得到 $1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28$ 共 $11$ 个不同整数。

所以正确答案是 C。

Listing $a/(15-a)$ for $1\le a\le14,$ the integers are $2,4,14;$ the fractions with fractional part $1/2$ are $\tfrac12,\tfrac32,\tfrac{13}{2};$ and $\tfrac14,\tfrac{11}{4}$ have complementary fractional parts $1/4,3/4.$

The remaining fractional parts are $1/14,2/13,4/11,2/3,7/8,1/7,$ and none of their complements occurs. Thus only the integer, half-integer, and quarter cases can give integer sums.

Integer pairs give $4, 6, 8, 16, 18, 28.$ Half-integer pairs $\left(\tfrac12, \tfrac32, \tfrac{13}{2}\right)$ give $1, 2, 3, 7, 8, 13.$ The quarter pair $\tfrac14 + \tfrac{11}{4}$ gives $3.$

The distinct integers are $1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28,$ a total of $11.$

Thus, the correct answer is C.

2021 AMC 12B Fall 第 5 题

答案：C

难度评级：1400

解答：

其他年份的第 5 题