Problem:
Let d(n) denote the number of positive integers that divide n, including 1 and n. For example, d(1)=1,d(2)=2, and d(12)=6. (This function is known as the divisor function.) Let
f(n)=3nβd(n)β
There is a unique positive integer N such that f(N)>f(n) for all positive integers nξ =N. What is the sum of the digits of N ?
Answer Choices:
A. 5
B. 6
C. 7
D. 8
E. 9 Solution:
Let n=βpβpΞ±pβ be the prime factorization of n. Then d(n)=βpβ(Ξ±pβ+1), so
For each prime p, the exponent Ξ± for which fpβ(Ξ±) is greatest must be found.
Note that fpβ(Ξ±+1)ξ =fpβ(Ξ±) because
(Ξ±+1Ξ±+2β)3ξ =p
Here the left-hand side is a decreasing function of Ξ± that tends to 1 as Ξ±ββ. Hence the set of Ξ± such that fpβ(Ξ±+1)>fpβ(Ξ±) is finite (and possibly empty). It follows that if
p>23=8,
then the optimal choice is Ξ±pβ=0. Similarly, if
8>p>(23β)3=827β
then the optimal choice is Ξ±pβ=1. If
827β>p>(34β)3=2764β
then the optimal choice is Ξ±pβ=2. Finally, if
2764β>p>(45β)3=64125β
then the optimal choice is Ξ±pβ=3. Note that (5) holds for all p>7, that (6) holds for p=5 and p=7, that (7) holds for p=3, and that (8) holds for p=2. Thus the extremal N is N=23β 32β 5β 7=2520. The requested sum of digits is 2+5+2+0=9β.