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音乐会成人比例变化

Adult Concert

专题
Discrete Math / 离散数学
难度
L4

题目详情

音乐会现场观众中,成人占 512\frac{5}{12}

一辆载有 50 人的大巴到达后,成人占比变为 1125\frac{11}{25}

问:大巴到达前,现场成人数量最少可能是多少?

512\frac{5}{12} of concert attendees are adults. A bus carrying 50 people arrives at the concert. Now, 1125\frac{11}{25} of concert attendees are adults. What is the minimum number of adults who could have been at the concert before the bus arrived?

解析

设大巴到达前总人数为 xx,则成人数为 512x\frac{5}{12}x,所以 x0(mod12)x\equiv 0\pmod{12}

到达后总人数为 x+50x+50,成人占比为 1125\frac{11}{25},因此 x+500(mod25)x+50\equiv 0\pmod{25}

500(mod25)50\equiv 0\pmod{25},所以等价于 x0(mod25)x\equiv 0\pmod{25}

同时满足 x0(mod12)x\equiv 0\pmod{12}x0(mod25)x\equiv 0\pmod{25} 的最小正解为 x=300x=300

于是原成人数最小为

512300=125.\frac{5}{12}\cdot 300=125.

Original Explanation

Let xx denote the total number of people at the concert before the bus arrives. x0mod12x \equiv 0 \mod 12. Then, x+500mod25x + 50 \equiv 0 \mod 25. Since 500mod2550 \equiv 0 \mod 25, we can rewrite the previous expression as x0mod25x \equiv 0 \mod 25. The minimum possible value of xx satisfying these two conditions is x=300x = 300. x512=125x \cdot \frac{5}{12} = 125.