AMC12 2018 B

AMC12 2018 B · Q17

AMC12 2018 B · Q17. It mainly tests Linear inequalities, Fractions.

Let $p$ and $q$ be positive integers such that $\frac{5}{9} < \frac{p}{q} < \frac{4}{7}$ and $q$ is as small as possible. What is $q - p$?

设 $p$ 和 $q$ 是正整数，使得 $\frac{5}{9} < \frac{p}{q} < \frac{4}{7}$ 且 $q$ 尽可能小。$q - p$ 是多少？

(A) 7 7

(B) 11 11

(D) 17 17

(E) 19 19

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The first inequality is equivalent to $9p>5q$, and because both sides are integers, it follows that $9p-5q\ge1$. Similarly, $4q-7p\ge1$. Now \[ \frac{1}{63}=\frac{4}{7}-\frac{5}{9}=\left(\frac{p}{q}-\frac{5}{9}\right)+\left(\frac{4}{7}-\frac{p}{q}\right) =\frac{9p-5q}{9q}+\frac{4q-7p}{7q} \ge\frac{1}{9q}+\frac{1}{7q} =\frac{16}{63q}. \] Thus $q\ge16$. Because \[ \frac{8}{16}<\frac{5}{9}<\frac{9}{16}<\frac{4}{7}<\frac{10}{16}, \] the fraction $\frac{9}{16}$ lies in the required interval, but $\frac{8}{16}$ and $\frac{10}{16}$ do not. Therefore when $q$ is as small as possible, $q=16$ and $p=9$, and the requested difference is $16-9=7$. Note: A theorem in the study of Farey fractions states that if $\frac{a}{p}<\frac{b}{q}$ and $bp-aq=1$, then the rational number with least denominator between $\frac{a}{p}$ and $\frac{b}{q}$ is $\frac{a+b}{p+q}$.

答案（A）：第一个不等式等价于 $9p>5q$，且因为两边都是整数，所以有 $9p-5q\ge1$。同理，$4q-7p\ge1$。现在 \[ \frac{1}{63}=\frac{4}{7}-\frac{5}{9}=\left(\frac{p}{q}-\frac{5}{9}\right)+\left(\frac{4}{7}-\frac{p}{q}\right) =\frac{9p-5q}{9q}+\frac{4q-7p}{7q} \ge\frac{1}{9q}+\frac{1}{7q} =\frac{16}{63q}. \] 因此 $q\ge16$。因为 \[ \frac{8}{16}<\frac{5}{9}<\frac{9}{16}<\frac{4}{7}<\frac{10}{16}, \] 分数 $\frac{9}{16}$ 落在所需区间内，但 $\frac{8}{16}$ 和 $\frac{10}{16}$ 不在其中。因此当 $q$ 尽可能小时，$q=16$ 且 $p=9$，所求差为 $16-9=7$。注：Farey 分数理论中的一个定理指出：若 $\frac{a}{p}<\frac{b}{q}$ 且 $bp-aq=1$，则位于 $\frac{a}{p}$ 与 $\frac{b}{q}$ 之间且分母最小的有理数为 $\frac{a+b}{p+q}$。

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