2021 Fall AMC 10B Problems/Problem 6
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
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_2
2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.