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=p1r1​​p2r2​​⋯pkrk​​, where the pj​ are distinct primes and the rj​ are positive integers. Then τ(110n3), the number of positive integer divisors of 110n3, is given by
τ(110n3)=(r1​+1)(r2​+1)⋯(rk​+1)=110
Because 110=2⋅5⋅11, it follows that k=3,{p1​,p2​,p3​}={2,5,11}, and, without loss of generality, r1​=1,r2​=4, and r3​=10. Therefore
n3=110p1​⋅p24​⋅p310​​=p23​⋅p39​, so n=p2​⋅p33​
It follows that 81n4=34⋅p24​⋅p312​, and because 3,p2​, and p3​ are distinct primes, τ(81n4)=5⋅5⋅13=325​.
The problems on this page are the property of the MAA's American Mathematics Competitions