Problem:
Let n be a positive integer greater than 4 such that the decimal representation of n! ends in k zeros and the decimal representation of (2n)! ends in 3k zeros. Let s denote the sum of the four least possible values of n. What is the sum of the digits of s?
Answer Choices:
A. 7
B. 8
C. 9
D. 10
E. 11
Solution:
Because there are ample factors of 2, it is enough to count the number of factors of 5. Let f(n) be the number of factors of 5 in positive integers less than or equal to n. For n from 5 to 9,f(n)=1. In order for f(2n) to equal 3,2n must be between 15 and 19, inclusive. Therefore n=8 or n=9. For n from 10 to 14,f(n)=2. In order for f(2n) to equal 6,2n must be between 25 and 29, inclusive. Hence, n=13 or n=14. Thus the four smallest integers n that satisfy the specified condition are 8,9,13, and 14. Their sum is 44 and the sum of the digits of 44 is (B)8.
OR
In fact there are only 4 possible values of n. By Legendre's Theorem, if n! ends in k zeros and (2n)! ends in k′ zeros, then
k=⌊5n⌋+⌊52n⌋+⌊53n⌋+⋯+⌊5jn⌋,
k′=⌊52n⌋+⌊522n⌋+⌊532n⌋+⋯+⌊5j2n⌋+⌊5j+12n⌋
where j is the highest power of 5 not exceeding n, and thus the highest power of 5 not exceeding 2n is at most j+1. If x is a real number, then ⌊2x⌋≤2⌊x⌋+1. So ⌊5i2n⌋≤2⌊5in⌋+1 for each 1≤i≤j+1. Adding these inequalities yields k′≤2k+j+1. If n≥15, then k>2+j−1=j+1 and k′<3k. For n=13 and n=14,k=2 and k′=5+1=6=3k. For n≤12,k=⌊5n⌋ and k′=⌊52n⌋; in this case k′=3k only for n=8 and n=9. So s=8+9+13+14=44 and the answer is 4+4=(B)8.
The problems on this page are the property of the MAA's American Mathematics Competitions