AMC10 2015 B

AMC10 2015 B · Q23

AMC10 2015 B · Q23. It mainly tests Primes & prime factorization, Number theory misc.

Let n be a positive integer greater than 4 such that the decimal representation of n! ends in k zeros and the decimal representation of (2n)! ends in 3k zeros. Let s denote the sum of the four least possible values of n. What is the sum of the digits of s ?

设$n$是一个大于4的正整数，使得$n!$的小数表示末尾有$k$个零，而$(2n)!$的小数表示末尾有$3k$个零。让$s$表示四个最小可能$n$值的总和。$s$的各位数字之和是多少？

(A) 7 7

(B) 8 8

(D) 10 10

(E) 11 11

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because there are ample factors of 2, it is enough to count the number of factors of 5. Let $f(n)$ be the number of factors of 5 in positive integers less than or equal to $n$. For $n$ from 5 to 9, $f(n)=1$. In order for $f(2n)$ to equal 3, $2n$ must be between 15 and 19, inclusive. Therefore $n=8$ or $n=9$. For $n$ from 10 to 14, $f(n)=2$. In order for $f(2n)$ to equal 6, $2n$ must be between 25 and 29, inclusive. Hence, $n=13$ or $n=14$. Thus the four smallest integers $n$ that satisfy the specified condition are 8, 9, 13, and 14. Their sum is 44 and the sum of the digits of 44 is 8.

答案（B）：因为 2 的因子数量充足，所以只需要统计 5 的因子数量。设 $f(n)$ 表示不超过 $n$ 的正整数中 5 的因子个数。对于 $n$ 从 5 到 9，$f(n)=1$。要使 $f(2n)=3$，则 $2n$ 必须在 15 到 19（含）之间。因此 $n=8$ 或 $n=9$。对于 $n$ 从 10 到 14，$f(n)=2$。要使 $f(2n)=6$，则 $2n$ 必须在 25 到 29（含）之间。因此 $n=13$ 或 $n=14$。所以满足给定条件的最小四个整数 $n$ 为 8、9、13 和 14。它们的和为 44，而 44 的各位数字和为 8。

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