# 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 1

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 $m+k = 16 + 42 = \boxed{\textbf{(B) }58}$

~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 $42 + 16 = \boxed{\textbf{(B) }58}$

~Arcticturn

## Solution 3

If a number has prime factorization $p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$, then the number of distinct positive divisors of this number is $\left( k_1 + 1 \right) \left( k_2 + 1 \right) \cdots \left( k_m + 1 \right)$.

We have $2021 = 43 \cdot 47$. Hence, if a number $N$ has 2021 distinct positive divisors, then $N$ takes one of the following forms: $p_1^{2020}$, $p_1^{42} p_2^{46}$.

Therefore, the smallest $N$ is $3^{42} 2^{46} = 2^4 \cdot 6^{42} = 16 \cdot 6^{42}$.

Therefore, the answer is $\boxed{\textbf{(B) }58}$.

~Steven Chen (www.professorchenedu.com)

## Video Solution

~Education, the Study of Everything

~savannahsolver

~IceMatrix