Problem:
For some positive integer n, the number 110n3 has 110 positive integer divisors, including 1 and the number 110n3. How many positive integer divisors does the number 81n4 have?
Answer Choices:
A. 110
B. 191
C. 261
D. 325
E. 425
Solution:
Let 110n3=p1r1ββp2r2βββ―pkrkββ, where the pjβ are distinct primes and the rjβ are positive integers. Then Ο(110n3), the number of positive integer divisors of 110n3, is given by
Ο(110n3)=(r1β+1)(r2β+1)β―(rkβ+1)=110
Because 110=2β
5β
11, it follows that k=3,{p1β,p2β,p3β}={2,5,11}, and, without loss of generality, r1β=1,r2β=4, and r3β=10. Therefore
n3=110p1ββ
p24ββ
p310ββ=p23ββ
p39β, so n=p2ββ
p33β
It follows that 81n4=34β
p24ββ
p312β, and because 3,p2β, and p3β are distinct primes, Ο(81n4)=5β
5β
13=325β.
The problems on this page are the property of the MAA's American Mathematics Competitions