Problem:
For each positive integer n, let
an=(n−1)!(n+9)!
Let k denote the smallest positive integer for which the rightmost nonzero digit of ak is odd. The rightmost nonzero digit of ak is
Answer Choices:
A. 1
B. 3
C. 5
D. 7
E. 9
Solution:
Factor an as a product of prime powers:
an=n(n+1)(n+2)⋯(n+9)=2e13e25e3⋯
Among the ten factors n,n+1,…,n+9, five are even and their product can be written 25m(m+1)(m+2)(m+3)(m+4). If m is even then m(m+2)(m+4) is divisible by 16 and thus e1≥9. If m is odd, then e1≥8. If e1>e3, then the rightmost nonzero digit of an is even. If e1≤e3, then the rightmost nonzero digit of an is odd. Hence we seek the smallest n for which e3≥e1. Among the ten numbers n,n+1,…,n+9, two are divisible by 5 and at most one of these is divisible by 25. Hence e3≥8 if and only if one of n,n+1,…,n+9 is divisible by 57. The smallest n for which an satisfies e3≥8 is thus n=57−9, but in this case the product of the five even numbers among n,n+1,…,n+9 is 25m(m+1)(m+2)(m+3)(m+4) where m is even, namely (57−9)/2=39058. As noted earlier, this gives e1≥9. For n=57−8=78117, the product of the five even numbers among n,n+1,…,n+9 is 25m(m+1)(m+2)(m+3)(m+4) with m=39059. Note that in this case e1=8. Indeed, 39059+1 is divisible by 4 but not by 8, and 39059+3 is divisible by 2 but not by 4. Compute the rightmost nonzero digit as follows. The odd numbers among n,n+1,…,n+9 are 78117,78119,78121,78123,78125=57 and the product of the even numbers 78118,78120,78122,78124,78126 is 25⋅39059⋅39060⋅39061⋅39062⋅39063=25⋅39059⋅(22⋅5⋅1953)⋅39061⋅(2⋅19531)⋅39063. (For convenience, we have underlined the needed unit digits.) Having written n(n+1)⋯(n+9) as 2858 times a product of odd factors not divisible by 5, we determine the rightmost nonzero digit by multiplying the units digits of these factors. It follows that, for n=57−8, the rightmost nonzero digit of an is the units digit of 7⋅9⋅1⋅3⋅9⋅3⋅1⋅1⋅3=(9⋅9)⋅(7⋅3)⋅(3⋅3), namely 9.