AMC10 2019 A

AMC10 2019 A · Q9

AMC10 2019 A · Q9. It mainly tests Primes & prime factorization, GCD & LCM.

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is not a divisor of the product of the first $n$ positive integers?

对于三位正整数 $n$，前 $n$ 个正整数的和不整除前 $n$ 个正整数的乘积的最大 $n$ 是多少？

(A) 995 995

(B) 996 996

(D) 998 998

(E) 999 999

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$, and the product of the first $n$ positive integers is $n!=n\cdot(n-1)\cdot\ldots\cdot2\cdot1$. If $n>1$ is odd, then $n\cdot\frac{n+1}{2}$ divides $n\cdot(n-1)!$ because $\frac{n+1}{2}$ is an integer between $1$ and $n$. If $n$ is even, then $\frac{n}{2}\cdot(n+1)$ does not divide $n!$ if and only if $n+1$ is prime. Because $997$ is the greatest three-digit prime number, the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is not a divisor of the product of the first $n$ positive integers is $997-1=996$.

答案（B）：前 $n$ 个正整数的和为 $\frac{n(n+1)}{2}$，前 $n$ 个正整数的积为 $n!=n\cdot(n-1)\cdot\ldots\cdot2\cdot1$。若 $n>1$ 为奇数，则 $n\cdot\frac{n+1}{2}$ 整除 $n\cdot(n-1)!$，因为 $\frac{n+1}{2}$ 是介于 $1$ 与 $n$ 之间的整数。若 $n$ 为偶数，则当且仅当 $n+1$ 为素数时，$\frac{n}{2}\cdot(n+1)$ 不整除 $n!$。由于 $997$ 是最大的三位素数，因此满足“前 $n$ 个正整数的和不是前 $n$ 个正整数的积的因数”的最大的三位正整数 $n$ 为 $997-1=996$。

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