Problem:
For each integer n≥4, let an denote the base- n number 0.133n. The product a4a5…a99 can be expressed as n!m, where m and n are positive integers and n is as small as possible. What is the value of m ?
Answer Choices:
A. 98
B. 101
C. 132
D. 798
E. 962
Solution:
Note that n3an=133⋅133n=an+n2+3n+3, so
an=n31n2+3n+3=n(n3−1)(n+1)3−1
Therefore
a4a5a99=4(43−1)53−15(53−1)63−199(993−1)10031=99!3!43−11003−1=99!66399(1002+100+1)=(21)(98!)(2)(10,101)=98!962
The problems on this page are the property of the MAA's American Mathematics Competitions