AMC10 2005 A

AMC10 2005 A · Q15

AMC10 2005 A · Q15. It mainly tests Primes & prime factorization, Counting divisors.

How many positive cubes divide $3! \cdot 5! \cdot 7!$?

有多少个正立方数整除 $3! \cdot 5! \cdot 7!$？

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) Written as a product of primes, we have $3!\cdot 5!\cdot 7!=2^8\cdot 3^4\cdot 5^2\cdot 7.$ A cube that is a factor has a prime factorization of the form $2^p\cdot 3^q\cdot 5^r\cdot 7^s$, where $p$, $q$, $r$, and $s$ are all multiples of $3$. There are $3$ possible values for $p$, which are $0$, $3$, and $6$. There are $2$ possible values for $q$, which are $0$ and $3$. The only value for $r$ and for $s$ is $0$. Hence there are $6=3\cdot 2\cdot 1\cdot 1$ distinct cubes that divide $3!\cdot 5!\cdot 7!$. They are $1=2^0 3^0 5^0 7^0,\quad 8=2^3 3^0 5^0 7^0,\quad 27=2^0 3^3 5^0 7^0,$ $64=2^6 3^0 5^0 7^0,\quad 216=2^3 3^3 5^0 7^0,\quad \text{and}\quad 1728=2^6 3^3 5^0 7^0.$

（E）写成素数的乘积，我们有 $3!\cdot 5!\cdot 7!=2^8\cdot 3^4\cdot 5^2\cdot 7.$ 作为因子的一个立方数，其素因数分解可写为 $2^p\cdot 3^q\cdot 5^r\cdot 7^s$，其中 $p,q,r,s$ 都是 $3$ 的倍数。$p$ 有 $3$ 个可能取值：$0,3,6$。$q$ 有 $2$ 个可能取值：$0,3$。$r$ 和 $s$ 的唯一取值都是 $0$。因此共有 $6=3\cdot 2\cdot 1\cdot 1$ 个不同的立方数能整除 $3!\cdot 5!\cdot 7!$。它们是 $1=2^0 3^0 5^0 7^0,\quad 8=2^3 3^0 5^0 7^0,\quad 27=2^0 3^3 5^0 7^0,$ $64=2^6 3^0 5^0 7^0,\quad 216=2^3 3^3 5^0 7^0,\quad \text{以及}\quad 1728=2^6 3^3 5^0 7^0.$

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