AMC10 2021 A

AMC10 2021 A · Q22

AMC10 2021 A · Q22. It mainly tests Averages (mean), Arithmetic misc.

Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$, the second sheet contains pages $3$ and $4$, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$. How many sheets were borrowed?

Hiram 的代数笔记有 50 页，印在 25 张纸上；第一张纸包含第 1 页和第 2 页，第二张纸包含第 3 页和第 4 页，依此类推。有一天他在去吃午饭前把笔记留在桌上，他的室友决定从笔记中间借一些页。当 Hiram 回来时，他发现室友拿走了一整套连续的纸张，并且剩下所有纸张上的页码平均数（均值）恰好为 19。借走了多少张纸？

(A) 10 10

(B) 13 13

(D) 17 17

(E) 20 20

Answer

Correct choice: (B)

正确答案：（B）

Solution

Suppose the roommate took sheets $a$ through $b$, or equivalently, page numbers $2a-1$ through $2b$. Because there are $(2b-2a+2)$ numbers taken, \[\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2}\] \[\implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.\] The first possible solution that comes to mind is if $2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12$, which indeed works, giving $b=22$ and $a=10$. The answer is $22-10+1=\boxed{\textbf{(B)} ~13}$.

假设室友拿走了从第 $a$ 张到第 $b$ 张的纸张，相当于页码从 $2a-1$ 到 $2b$。被拿走的页数有 $(2b-2a+2)$ 个， \[\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2}\] \[\implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.\] 第一个可能的解是 $2a+2b-39=25$，$b-a+1=13 \implies a+b=32$，$b-a=12$，这确实成立，给出 $b=22$ 和 $a=10$。答案为 $22-10+1=\boxed{\textbf{(B)} ~13}$。

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