2021 Fall AMC 10B Problems/Problem 6: Difference between revisions
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==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}$ | |||
~Arcticturn | |||
==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}} | ||
Revision as of 10:52, 23 November 2021
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
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 
p_1
3
p_22
2^4
2^42 \cdot 3^42
m
2^4
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. 
