2021 Fall AMC 10B Problems/Problem 6
Contents
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 Education
Video Solution by WhyMath
~savannahsolver
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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