Difference between revisions of "2021 Fall AMC 10B Problems/Problem 6"
(→Video Solution by Interstigation) |
m |
||
(11 intermediate revisions by 5 users not shown) | |||
Line 8: | Line 8: | ||
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 | + | 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 | ||
Line 29: | Line 29: | ||
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
− | ==Video Solution | + | ==Video Solution by Interstigation== |
+ | https://youtu.be/p9_RH4s-kBA?t=530 | ||
+ | |||
+ | ==Video Solution== | ||
https://youtu.be/bRohyPen8ik | https://youtu.be/bRohyPen8ik | ||
− | ~Education, the Study of | + | ~Education, the Study of Everything |
− | |||
− | |||
− | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/bErxcXXwWkw | ||
− | ==Video Solution by | + | ~savannahsolver |
− | https://youtu.be/ | + | ==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
Contents
[hide]Problem
The least positive integer with exactly distinct positive divisors can be written in the form , where and are integers and is not a divisor of . What is
Solution 1
Let this positive integer be written as . The number of factors of this number is therefore , and this must equal 2021. The prime factorization of 2021 is , so and . To minimize this integer, we set and . Then this integer is . Now and so
~KingRavi
Solution 2
Recall that can be written as . Since we want the integer to have divisors, we must have it in the form , where and are prime numbers. Therefore, we want to be and to be . To make up the remaining , we multiply by , which is which is . Therefore, we have
~Arcticturn
Solution 3
If a number has prime factorization , then the number of distinct positive divisors of this number is .
We have . Hence, if a number has 2021 distinct positive divisors, then takes one of the following forms: , .
Therefore, the smallest is .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=530
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.