2021 Fall AMC 10B Problems/Problem 6

Problem

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 $m+k?$

$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$

Solution

Let this positive integer be written as $p_1^{e_1}\cdot p_2^{e_2}$ . The number of factors of this number is therefore $(e_1+1) \cdot (e_2+1)$ , and this must equal 2021. The prime factorization of 2021 is $43 \cdot 47$ , so $e_1+1 = 43 \implies e_1=42$ and $e_2+1=47\implies e_2=46$ . To minimize this integer, we set $p_1 = 3$ and $p_2 = 2$ . Then this integer is $3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}$ . Now $m=16$ and $k=42$ so $m+k = 16 + 42 = 58 = \boxed{B}$

~KingRavi

Solution 2

Recall that $6^k$ can be written as $2^k \cdot 3^k$ . Since we want the integer to have $2021$ divisors, we must have it in the form $p_1^42 \cdot p_2^46$ , where $p_1$ and $p_2$ are prime numbers. Therefore, we want $p_1$ to be $3$ and $p_2$ to be $2$ . To make up the remaining $2^4$ , we multiply $2^42 \cdot 3^42$ by $m$ , which is $2^4$ which is 16. Therefore, we have $42 + 16 = \boxed {(B) 58}$

~Arcticturn

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2021 Fall AMC 10B Problems/Problem 6

Contents

Problem

Solution

Solution 2

See Also