Question

恰有 $2021$ 个不同正因数的最小正整数可写成 $m \cdot 6^k$ ，其中 $m$ 和 $k$ 为整数，且 $6$ 不是 $m$ 的因数。求 $m+k$ 。

The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k,$ where $m$ and $k$ are integers and $6$ is not a divisor of $m.$ What is $m+k?$

$47$

$58$

$59$

$88$

$90$

答案：B

知识点：因数个数质因数分解最优化

难度评级：1420

解答：

$z$ $(e_1+1)(e_2+1) \cdots$ $z=p_1^{e_1} p_2^{e_2} \cdots,$

若一个数 $2021$ 的质因数分解为则其正因数个数为 $2021 = 43\cdot 47$ 。 $p_1^{46}p_2^{42}$ $p^{2020}$ 。 $2021=(e_1+1)(e_2+1) \cdots,$

若所求数有 $p_1 = 2,p_2=3$ 个因数，则因为，所以该数必为或。 $2^{46}3^{42} = 16\cdot 6^{42} .$

最小数来自第一种形式，并取，即让较小质数承载较大指数： $m=16$ $k=42,$ $m+k = 42+16 = 58.$

所以答案是 B。

Before starting, note that if we can represent the prime factorization of an integer $z$ as $z=p_1^{e_1} p_2^{e_2} \cdots,$ then there are $(e_1+1)(e_2+1) \cdots$ distinct positive factors.

If the number in question has $2021$ factors, by the previous logic, $2021=(e_1+1)(e_2+1) \cdots,$ and as the prime factorization of $2021 = 43\cdot 47,$ then our number must be $p_1^{46}p_2^{42}$ or $p^{2020}.$

The smallest number we can make in either of these is making $p_1 = 2,p_2=3$ in the first configuration, yielding $2^{46}3^{42} = 16\cdot 6^{42} .$

Therefore, $m=16$ $k=42,$ so $m+k = 42+16 = 58.$

Thus, the answer is B .

2021 AMC 10B Fall 第 6 题

答案：B

难度评级：1420

解答：

其他年份的第 6 题