AMC10 2019 B

AMC10 2019 B · Q7

AMC10 2019 B · Q7. It mainly tests Primes & prime factorization, GCD & LCM.

Each piece of candy in a shop costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs 20 cents. What is the least possible value of $n$?

商店里的每块糖果的价格都是整数量分的。Casper 的钱恰好够买 12 块红糖果、14 块绿糖果、15 块蓝糖果，或 $n$ 块紫糖果。一块紫糖果的价格是 20 分。$n$ 的最小可能值是多少？

(A) 18 18

(B) 21 21

(D) 25 25

(E) 28 28

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because Casper has exactly enough money to buy 12 pieces of red candy, the amount of money he has must be a multiple of 12 cents. Similarly, it must be a multiple of both 14 cents and 15 cents. Furthermore, this amount of money will buy a whole number of purple candies that cost 20 cents each, so the amount of money must also be a multiple of 20. The least common multiple of $12=2^2\cdot 3$, $14=2\cdot 7$, $15=3\cdot 5$, and $20=2^2\cdot 5$ is $2^2\cdot 3\cdot 5\cdot 7=420$. Therefore the number of purple candies that Casper can buy, $n$, must be a multiple of $420\div 20=21$. Thus the least possible value of $n$ is 21. In this case the red candies cost $420\div 12=35$ cents each, the green candies cost $420\div 14=30$ cents each, and the blue candies cost $420\div 15=28$ cents each.

答案（B）：因为 Casper 的钱恰好够买 12 颗红色糖果，所以他拥有的钱数必须是 12 美分的倍数。同样，它也必须同时是 14 美分和 15 美分的倍数。此外，这笔钱还必须能买到整数颗每颗 20 美分的紫色糖果，因此金额也必须是 20 的倍数。$12=2^2\cdot 3$、$14=2\cdot 7$、$15=3\cdot 5$、$20=2^2\cdot 5$ 的最小公倍数是 $2^2\cdot 3\cdot 5\cdot 7=420$。因此，Casper 能买的紫色糖果数量 $n$ 必须是 $420\div 20=21$ 的倍数。所以 $n$ 的最小可能值是 21。在这种情况下，红色糖果每颗 $420\div 12=35$ 美分，绿色糖果每颗 $420\div 14=30$ 美分，蓝色糖果每颗 $420\div 15=28$ 美分。

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