AMC12 2010 B

AMC12 2010 B · Q25

AMC12 2010 B · Q25. It mainly tests Primes & prime factorization, Counting divisors.

For every integer $n\ge2$, let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$. For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$. What is the largest integer $m$ such that $2010^m$ divides $\prod_{n=2}^{5300}\text{pow}(n)$?

对每个整数 $n\ge2$，令 $\text{pow}(n)$ 表示 $n$ 的最大素因子的最大幂。例如 $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$。求最大的整数 $m$，使得 $2010^m$ 整除 $\prod_{n=2}^{5300}\text{pow}(n)$。

(A) 74 74

(B) 75 75

(D) 77 77

(E) 78 78

Answer

Correct choice: (D)

正确答案：（D）

Solution

Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$, there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times $67$ is incorporated into the giant product. All numbers $n=67 \cdot x$, given $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$ such that for any integer $x$ between $1$ and $m$, prime $p_x$ must be less than $67$, contributes a 67 to the product. Considering $67 \cdot 79 < 5300 < 67 \cdot 80$, the possible values of x are $1,2,...,70,72,74,...78$, since $x=71,73,79$ are primes that are greater than 67. However, $\text{pow}\left(67^2\right)$ contributes two $67$s to the product, so we must count it twice. Therefore, the answer is $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$.

因为 $67$ 是 $2010$ 的最大素因子，这意味着在 $\prod_{n=2}^{5300}\text{pow}(n)$ 的素因数分解中，会出现形如 $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ 的项，其中 $x$ 就是我们要找的值。因此，为求答案，我们需要统计在这个巨大乘积中 $67$ 出现了多少次。所有形如 $n=67 \cdot x$ 的数，其中 $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$，并且对任意整数 $x$ 在 $1$ 到 $m$ 之间，素数 $p_x$ 必须小于 $67$，都会向乘积贡献一个 $67$。由于 $67 \cdot 79 < 5300 < 67 \cdot 80$，$x$ 的可能取值为 $1,2,...,70,72,74,...78$，因为 $x=71,73,79$ 是大于 $67$ 的素数。然而，$\text{pow}\left(67^2\right)$ 会向乘积贡献两个 $67$，所以必须将其计数两次。因此答案为 $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$。

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