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}

Page

Toolbox

Search

2021 Fall AMC 10B Problems/Problem 6

Problem

Solution