AMC12 2003 A

AMC12 2003 A · Q23

AMC12 2003 A · Q23. It mainly tests Primes & prime factorization, Counting divisors.

How many perfect squares are divisors of the product $1!\cdot 2!\cdot 3!\cdots 9!$?

乘积 $1!\cdot 2!\cdot 3!\cdots 9!$ 的因数中，有多少个是完全平方数？

(A) 504 504

(B) 672 672

(D) 936 936

(E) 1008 1008

Answer

Correct choice: (B)

正确答案：（B）

Solution

We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$. To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{\mathrm{(B)}}$.

我们要找从 $1 - 9$ 的所有阶乘之积中，完全平方因数的个数。可以将其展开并提取阶乘，然后对整个乘积做质因数分解；也可以通过统计每个质因数在各个阶乘中出现的次数来得到该分解。结果为 $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$。要成为完全平方数，其质因数分解中的每个指数必须为偶数：$2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$。为求总的可能数，对每个指数加 $1$ 后相乘，得到 $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{\mathrm{(B)}}$。

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