The number of digits of a number is given by [log(x)]+1
where [ ] denotes the greatest integer and the logarithm is base 10.
So we are trying to show [log(4
^{n})]+1 +[log(25
^{n})]+1 is odd
which is equivalent to showing [log(4
^{n})]+[log(25
^{n})] is odd
[log(4
^{n})]+[log(25
^{n})] = [n*log(4)]+[n*log(25)]
= [n*(2log(25))]+[n*log(25)]
= [2n  n*log(25)]+[n*log(25)]
What I now need to show is for any integer a and noninteger b
[ab]+[b] = a1
which it is because
[ab]=a+[b]
and
[b]+[b]=1
So the above expression becomes 2n1

Posted by Jer
on 20140429 14:10:04 