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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 6"

(Created page with "==Problem== The least positive integer with exactly <math>2021</math> distinct positive divisors can be written in the form <math>m \cdot 6^k</math>, where <math>m</math> and...")
 
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==Solution==
 
==Solution==
  
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Let this positive integer be written as <math>p_1^{e_1}\cdot p_2^{e_2}</math>. The number of factors of this number is therefore <math>(e_1+1) \cdot (e_2+1)</math>, and this must equal 2021. The prime factorization of 2021 is <math>43 \cdot 47</math>, so <math>e_1+1 = 43 \implies e_1=42</math> and <math>e_2+1=47\implies e_2=46</math>. To minimize this integer, we set <math>p_1 = 3</math> and <math>p_2 = 2</math>. Then this integer is 3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}<math>.
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Let this positive integer be written as <math>p_1^{e_1}\cdot p_2^{e_2}</math>. The number of factors of this number is therefore <math>(e_1+1) \cdot (e_2+1)</math>, and this must equal 2021. The prime factorization of 2021 is <math>43 \cdot 47</math>, so <math>e_1+1 = 43 \implies e_1=42</math> and <math>e_2+1=47\implies e_2=46</math>. To minimize this integer, we set <math>p_1 = 3</math> and <math>p_2 = 2</math>. Then this integer is <math>3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}</math>.
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Now </math>m=16<math> and </math>k=42<math> so </math>m+k = 16 + 42 = 58 = \boxed{B}
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Now <math>m=16</math> and <math>k=42</math> so <math>m+k = 16 + 42 = 58 = \boxed{B}</math>
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~KingRavi

Revision as of 00:42, 23 November 2021

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

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