Problem:
Call a positive integer n k-pretty if n has exactly k positive divisors and n is divisible by k. For example, 18 is 6-pretty. Let S be the sum of positive integers less than 2019 that are 20-pretty. Find 20S.
Solution:
We begin by observing that any 20-pretty integer must be divisible by both 4 and 5, meaning it includes at least 22 and 5 in its prime factorization. Thus, let n=2a⋅5b⋅k, where a≥2, b≥1, and gcd(k,10)=1. The number of divisors of n is given by d(n)=(a+1)(b+1)d(k)=20.
We examine possible factorizations of 20 into (a+1)(b+1)d(k) under the constraints that a+1≥3 and b+1≥2.
Case 1: (a+1,b+1)=(5,2)⇒(a,b)=(4,1)
Then d(k)=1020=2. So k must be a prime, and since gcd(k,10)=1, we only consider primes not divisible by 2 or 5. Also, we require n=24⋅51⋅k=80k<2020⇒k<25.25.
Valid primes k are 3,7,11,13,17,19,23. The corresponding values of n are:
80⋅(3+7+11+13+17+19+23)=80⋅93=7440
Case 2: (a+1,b+1)=(5,4)⇒(a,b)=(4,3)
Then d(k)=2020=1, so k=1. This gives:
n=24⋅53=16⋅125=2000<2020
Other Cases: For instance, (a+1,b+1)=(4,5) or (10,2) result in n>2020 and are excluded.
Total sum: S=7440+2000=9440, so
20S=209440=472
The problems on this page are the property of the MAA's American Mathematics Competitions