AMC10 2001 A

AMC10 2001 A · Q12

AMC10 2001 A · Q12. It mainly tests Divisibility & factors, Parity (odd/even).

Suppose that n is the product of three consecutive integers and that n is divisible by 7. Which of the following is not necessarily a divisor of n?

假设 $n$ 是三个连续整数的乘积，并且 $n$ 能被 7 整除。以下哪一个不一定是 $n$ 的因数？

(A) 6 6

(B) 14 14

(D) 28 28

(E) 42 42

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) In any triple of consecutive integers, at least one is even and one is a multiple of 3. Therefore, the product of the three integers is both even and a multiple of 3. Since 7 is a divisor of the product, the numbers 6, 14, 21, and 42 must also be divisors of the product. However, 28 contains two factors of 2, and $n$ need not. For example, $5 \cdot 6 \cdot 7$ is divisible by 7, but not by 28.

（D）在任意三个连续整数中，至少有一个是偶数，且至少有一个是 3 的倍数。因此，这三个整数的乘积既是偶数又是 3 的倍数。由于 7 是该乘积的一个因数，那么 6、14、21 和 42 也必定都是该乘积的因数。然而，28 含有两个 2 的因子，而 $n$ 未必如此。例如，$5 \cdot 6 \cdot 7$ 能被 7 整除，但不能被 28 整除。

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