AMC10 2020 A

AMC10 2020 A · Q15

AMC10 2020 A · Q15. It mainly tests Probability (basic), Primes & prime factorization.

A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

随机选择 $12!$ 的一个正整数除数。所选除数是完全平方的概率可以表示为 $\frac{m}{n}$，其中 $m$ 和 $n$ 是互质的正整数。求 $m+n$？

(A) 3 3

(B) 5 5

(D) 18 18

(E) 23 23

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The prime factorization of 12! = 12 · 11 · · · 1 is $2^{10}\cdot 3^{5}\cdot 5^{2}\cdot 7\cdot 11$. To choose one divisor at random is equivalent to choosing an exponent for 2 at random from $\{0,1,2,\ldots,10\}$, an exponent for 3 at random from $\{0,1,2,\ldots,5\}$, an exponent for 5 at random from $\{0,1,2\}$, and exponents for 7 and 11 at random from $\{0,1\}$. It follows that there are $(10+1)\cdot(5+1)\cdot(2+1)\cdot(1+1)\cdot(1+1)=11\cdot 6\cdot 3\cdot 2\cdot 2$ possible divisors. The chosen divisor will be a perfect square if and only if all the chosen exponents are even. There are 6 even choices for the exponent of 2 (namely 0, 2, 4, 6, 8, or 10), 3 even choices for the exponent of 3, 2 even choices for the exponent of 5, and 1 even choice for the exponent of 7 and the exponent of 11. The probability that the chosen divisor is a perfect square is therefore $\dfrac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\dfrac{1}{22}$. The requested sum is $1+22=23$.

答案（E）：$12!=12\cdot 11\cdot\cdots\cdot 1$ 的质因数分解为 $2^{10}\cdot 3^{5}\cdot 5^{2}\cdot 7\cdot 11$。随机选取一个因数，等价于：从 $\{0,1,2,\ldots,10\}$ 中随机选取 2 的指数；从 $\{0,1,2,\ldots,5\}$ 中随机选取 3 的指数；从 $\{0,1,2\}$ 中随机选取 5 的指数；并从 $\{0,1\}$ 中随机选取 7 和 11 的指数。因此因数总数为 $(10+1)\cdot(5+1)\cdot(2+1)\cdot(1+1)\cdot(1+1)=11\cdot 6\cdot 3\cdot 2\cdot 2$。所选因数是完全平方数当且仅当所有指数都是偶数。2 的指数有 6 种偶数选法（即 0、2、4、6、8、10），3 的指数有 3 种偶数选法，5 的指数有 2 种偶数选法，而 7 与 11 的指数各只有 1 种偶数选法。因此所选因数为完全平方数的概率为 $\dfrac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\dfrac{1}{22}$。所求的和为 $1+22=23$。

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