Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

← Older editNewer edit →
Arcticturn (talk | contribs)
editor
651 edits
Arcticturn (talk | contribs)
editor
651 edits
Line 13: Line 13:


==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 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 16. Therefore, we have </math>42 + 16 = \boxed {(B) 58}$
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 16. Therefore, we have <math>42 + 16 = \boxed {(B) 58}</math>


~Arcticturn
~Arcticturn

Revision as of 10:52, 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

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

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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_6&oldid=165429"