AMC12 2007 A

AMC12 2007 A · Q14

AMC12 2007 A · Q14. It mainly tests Word problems (algebra), Primes & prime factorization.

Let $a$, $b$, $c$, $d$, and $e$ be distinct integers such that $(6-a)(6-b)(6-c)(6-d)(6-e)=45$ What is $a+b+c+d+e$?

设 $a$, $b$, $c$, $d$, 和 $e$ 为互不相同的整数，使得 $(6-a)(6-b)(6-c)(6-d)(6-e)=45$ 求 $a+b+c+d+e$。

(A) 5 5

(B) 17 17

(D) 27 27

(E) 30 30

Answer

Correct choice: (C)

正确答案：（C）

Solution

If $45$ is expressed as a product of five distinct integer factors, the absolute value of the product of any four is at least $|(-3)(-1)(1)(3)|=9$, so no factor can have an absolute value greater than $5$. Thus the factors of the given expression are five of the integers $\pm 3, \pm 1, \pm 5$. The product of all six of these is $-225=(-5)(45)$, so the factors are $-3, -1, 1, 3,$ and $5.$ The corresponding values of $a, b, c, d,$ and $e$ are $9, 7, 5, 3,$ and $1,$ and their sum is $\fbox{25 (C)}$

若将 $45$ 表示为五个互不相同的整数因子的乘积，则任意四个因子的乘积的绝对值至少为 $|(-3)(-1)(1)(3)|=9$，因此没有因子的绝对值能大于 $5$。所以该表达式的因子是 $\pm 3, \pm 1, \pm 5$ 这六个整数中的五个。六个数的乘积为 $-225=(-5)(45)$，因此这五个因子为 $-3, -1, 1, 3,$ 和 $5.$ 对应的 $a, b, c, d,$ 和 $e$ 的值为 $9, 7, 5, 3,$ 和 $1,$ 它们的和为 $\fbox{25 (C)}$

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