AMC10 2001 A

AMC10 2001 A · Q14

AMC10 2001 A · Q14. It mainly tests Linear equations, Divisibility & factors.

A charity sells 140 benefit tickets for a total of $2001. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?

一个慈善机构售出了 140 张总价为 $2001 的福利票。有些票按全价（整美元金额）出售，其余按半价出售。全价票筹集了多少钱？

(A) $782$ $782$

(B) $986$ $986$

(D) $1219$ $1219$

(E) $1449$ $1449$

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) Let $n$ be the number of full-price tickets and $p$ be the price of each in dollars. Then $np+(140-n)\cdot\frac{p}{2}=2001,$ so $p(n+140)=4002.$ Thus $n+140$ must be a factor of $4002=2\cdot3\cdot23\cdot29.$ Since $0\le n\le140,$ we have $140\le n+140\le280,$ and the only factor of $4002$ that is in the required range for $n+140$ is $174=2\cdot3\cdot29.$ Therefore, $n+140=174,$ so $n=34$ and $p=23.$ The money raised by the full-price tickets is $34\cdot23=782$ dollars.

（A）设 $n$ 为全价票的数量，$p$ 为每张票的价格（美元）。则 $np+(140-n)\cdot\frac{p}{2}=2001,$ 所以 $p(n+140)=4002.$ 因此 $n+140$ 必须是 $4002=2\cdot3\cdot23\cdot29$ 的一个因数。由于 $0\le n\le140,$ 有 $140\le n+140\le280,$ 而 $4002$ 在 $n+140$ 所需范围内的唯一因数是 $174=2\cdot3\cdot29.$ 因此 $n+140=174,$ 所以 $n=34,$ $p=23.$ 全价票筹集到的金额为 $34\cdot23=782$ 美元。

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