Problem:
Let Ο(n) denote the number of positive integer divisors of n. Find the sum of the six least positive integers n that are solutions to Ο(n)+Ο(n+1)=7.
Solution:
Let p,q, and r represent primes. Because Ο(n)=1 only for n=1, there is no n for which {Ο(n),Ο(n+1)}={1,6}. If {Ο(n),Ο(n+1)}={2,5}, then {n,n+1}={p,q4}, so β£β£β£βpβq4β£β£β£β=1. Checking q=2 and p=17 yields the solution n=16. If q>2, then q is odd, and p=q4Β±1 is even, so p cannot be prime.\
If {Ο(n),Ο(n+1)}={3,4}, then {n,n+1}={p2,q3} or {p2,qr}. Consider β£β£β£βp2βq3β£β£β£β=1. If p2β1=(pβ1)(p+1)=q3, then q=2. This yields the solution p=3 and q=2, so n=8. If q3β1=(qβ1)(q2+q+1)=p2, then qβ1=1, which does not give a solution. Consider β£β£β£βp2βqrβ£β£β£β=1. If p2β1=(pβ1)(p+1)=qr, then if p>2, the left side is divisible by 8 , so there are no solutions. Finding the smallest four primes such that p2+1=qr gives 32+1=10,52+1=26,112+1=122, and 192+1=362. The six least values of n are 8,9,16,25,121, and 361 , whose sum is 540 .
The problems on this page are the property of the MAA's American Mathematics Competitions