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

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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>.
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>.
Now <math>m=16</math> and <math>k=42</math> so <math>m+k = 16 + 42 = 58 = \boxed{B}</math>
Now <math>m=16</math> and <math>k=42</math> so <math>m+k = 16 + 42 = \boxed{\textbf{(B) }58}</math>


~KingRavi
~KingRavi


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


~Arcticturn
~Arcticturn
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https://youtu.be/p9_RH4s-kBA?t=530
https://youtu.be/p9_RH4s-kBA?t=530


==Video Solution==
https://youtu.be/bRohyPen8ik
~Education, the Study of Everything
==Video Solution by WhyMath==
https://youtu.be/bErxcXXwWkw
~savannahsolver
==Video Solution by TheBeautyofMath==
https://youtu.be/RyN-fKNtd3A
~IceMatrix
==See Also==
==See Also==
{{AMC10 box|year=2021 Fall|ab=B|num-a=7|num-b=5}}
{{AMC10 box|year=2021 Fall|ab=B|num-a=7|num-b=5}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 23:55, 29 December 2022

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 by Interstigation

https://youtu.be/p9_RH4s-kBA?t=530

Video Solution

https://youtu.be/bRohyPen8ik

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/bErxcXXwWkw

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/RyN-fKNtd3A

~IceMatrix

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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