AMC10 2015 A

AMC10 2015 A · Q15

AMC10 2015 A · Q15. It mainly tests Linear equations, GCD & LCM.

Consider the set of all fractions $\frac{x}{y}$, where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$, the value of the fraction is increased by $10\%$?

考虑所有形如$\frac{x}{y}$的分数集合，其中$x$和$y$是互质的正整数。有多少个这样的分数满足：如果分子和分母都增加$1$，那么该分数的值增加了$10\%$？

(A) 0 0

(B) 1 1

(D) 3 3

(E) infinitely many 无数

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$, it follows that $10y-11x-xy=0$ and so $(10-x)(11+y)=110=2\cdot5\cdot11$. The only possible values of $10-x$ are $5$, $2$, and $1$ because $x$ and $y$ are positive integers. Thus the possible values of $x$ are $5$, $8$, and $9$. Of the resulting fractions $\frac{5}{11}$, $\frac{8}{44}$, and $\frac{9}{99}$, only the first is in simplest terms.

答案（B）：因为 $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$，可得 $10y-11x-xy=0$，因此 $(10-x)(11+y)=110=2\cdot5\cdot11$。由于 $x$ 和 $y$ 是正整数，$10-x$ 的可能取值只有 $5$、$2$ 和 $1$。因此 $x$ 的可能取值为 $5$、$8$ 和 $9$。由此得到的分数 $\frac{5}{11}$、$\frac{8}{44}$ 和 $\frac{9}{99}$ 中，只有第一个是最简分数。

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