The solution can easily be found by trying powers. For example with a base of 6. 6^4 has 4 digits but 6^5 falls short.
What is then sought is for b^n > 10^(n1)
nlog(b)=n1
log(b)=11/n
1/n = 1log(b)
n=1/(1log(b))
for b=9 this gives 21.85 implying 21 is the largest n.
b largest n
9 21
8 10
7 6
6 4
5 3
4 2
3 1
2 1
1 1
The sum of the second column is 49.

Posted by Jer
on 20151020 13:42:48 