Problem:
The sum of the base- 10 logarithms of the divisors of 10n is 792 . What is n ?
Answer Choices:
A. 11
B. 12
C. 13
D. 14
E. 15
Solution:
Because the prime factorization of 10 is 2⋅5, the positive divisors of 10n are the numbers 2a⋅5b with 0≤a≤n and 0≤b≤n. Thus
792=a=0∑nb=0∑nlog10(2a5b)=a=0∑nb=0∑n(alog102+blog105)=b=0∑na=0∑n(alog102)+a=0∑nb=0∑n(blog105)=(n+1)(log102)a=0∑na+(n+1)(log105)b=0∑nb=(n+1)(log102+log105)(21n(n+1))=21n(n+1)2(log1010)=21n(n+1)2
Hence n(n+1)2=2⋅792=2⋅11⋅72=11⋅122, so n=11.
OR
Let d(M) denote the number of divisors of a positive integer M. The sum of the logs of the divisors of M is equal to the log of the product of its divisors.
If M is not a square, its divisors can be arranged in pairs, each with a product of M. Thus the product of the divisors is Md(M)/2. A similar argument shows that this result is also true if M is a square. Therefore
792=log((10n)d(10n)/2)=21d(10n)⋅n=21d(2n⋅5n)⋅n=21(n+1)2⋅n
and the conclusion follows as in the first solution.
The problems on this page are the property of the MAA's American Mathematics Competitions