Problem:
The least positive integer with exactly 2021 distinct positive divisors can be written in the form mâ‹…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 p1e1​​⋅p2e2​​. The number of factors of this number is therefore (e1​+1)⋅(e2​+1). and this must equal 2021. The prime factorization of 2021 is 43⋅47, so e1​+1=43⇒e1​=42 and e2​+1=47⇒e2​=46. To minimize the integer, we set p1​=3 and p2​=2. Then the integer is 342⋅246=24⋅242⋅342=16⋅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