AMC10 2020 A

AMC10 2020 A · Q9

AMC10 2020 A · Q9. It mainly tests Fractions, GCD & LCM.

A single bench section at a school event can hold either 7 adults or 11 children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N$?

学校活动中的一个长椅段可以容纳7个成人或11个儿童。当$N$个长椅段首尾相连时，相同数量的成人和儿童坐在一起将正好占满所有长椅空间。$N$的最小正整数值是多少？

(A) 9 9

(B) 18 18

(D) 36 36

(E) 77 77

Answer

Correct choice: (B)

正确答案：（B）

Solution

One adult and one child will occupy $\frac{1}{7} + \frac{1}{11} = \frac{18}{77}$ of a bench section. The problem is asking for the least possible positive integer value of $N$ that will make $\frac{77}{18} \cdot N$ an integer. Because 77 and 18 are relatively prime, the least such value is $N = 18$, in which case 77 adults and 77 children will occupy all the bench space.

一个成人和一个儿童将占用$\frac{1}{7} + \frac{1}{11} = \frac{18}{77}$个长椅段的空间。问题要求$N$的最小正整数值，使得$\frac{77}{18} \cdot N$为整数。因为77和18互质，最小值为$N = 18$，此时77个成人和77个儿童将正好占满所有长椅空间。

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