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...") |
m |
||
(21 intermediate revisions by 10 users not shown) | |||
Line 5: | Line 5: | ||
<math>(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90</math> | <math>(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | 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>. | + | 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 < | + | Now <math>m=16</math> and <math>k=42</math> so <math>m+k = 16 + 42 = \boxed{\textbf{(B) }58}</math> |
+ | |||
+ | ~KingRavi | ||
+ | |||
+ | ==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{\textbf{(B) }58}</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Solution 3 == | ||
+ | If a number has prime factorization <math>p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}</math>, then the number of distinct positive divisors of this number is <math>\left( k_1 + 1 \right) \left( k_2 + 1 \right) \cdots \left( k_m + 1 \right)</math>. | ||
+ | |||
+ | We have <math>2021 = 43 \cdot 47</math>. | ||
+ | Hence, if a number <math>N</math> has 2021 distinct positive divisors, then <math>N</math> takes one of the following forms: <math>p_1^{2020}</math>, <math>p_1^{42} p_2^{46}</math>. | ||
+ | |||
+ | Therefore, the smallest <math>N</math> is <math>3^{42} 2^{46} = 2^4 \cdot 6^{42} = 16 \cdot 6^{42}</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) }58}</math>. | ||
+ | |||
+ | ~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== | ||
+ | {{AMC10 box|year=2021 Fall|ab=B|num-a=7|num-b=5}} | ||
+ | {{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.