Problem:
The least positive integer with exactly 2021 distinct positive divisors can be written in the form mβ
6k, where m and k are integers and 6 is not a divisor of m. What is m+k?
Answer Choices:
A. 47
B. 58
C. 59
D. 88
E. 90
Solution:
Let this positive integer be written as p1e1βββ
p2e2ββ. The number of factors of this number is therefore (e1β+1)β
(e2β+1). and this must equal 2021. The prime factorization of 2021 is 43β
47, so e1β+1=43βe1β=42 and e2β+1=47βe2β=46. To minimize the integer, we set p1β=3 and p2β=2. Then the integer is 342β
246=24β
242β
342=16β
642. Now m=16 and k=42 so m+k = 16+42 = \boxed
The problems on this page are the property of the MAA's American Mathematics Competitions