*If you are in Hong Kong and you need help for university mathematics courses, please visit www.all-r-math.com.*

(Easter Holiday gives me another excuse to be lazy....)

A simple number theory question: Show that there is at least one (and hence infinitely many) multiples of

*p*among 1

_{b}, 11

_{b}, 111

_{b}, 1111

_{b}, 11111

_{b},... for all but finitely many prime numbers

*p*. Here 11111

_{b}etc are numbers written with base b numerals.

Answer after the good-looking easter eggs :P

Let n(k)=111...1

_{b}denote the number with k 1's. Note that

n(k)=$b^{k-1}+b^{k-2}+b^{k-3}+\cdots+1$=(b

^{k}-1)/(b-1).

We claim that when

*p*is not a factor of b, then n(k) is a multiple of

*p*for some k.

Use factor theorem, we know that $x^{k-1}+x^{k-2}+x^{k-3}+\cdots+1=(x-1)q(x)+k$. Consequently, n(k)=k mod (b-1). Therefore if

*p*is a factor of b-1 (and hence not a factor of b), then

*p*divides n(b-1).

By Fermat's little theorem, we know that if

*p*is not a factor of b, then

*p*divides b

^{p-1}-1. Therefore if

*p*is neither a factor of b nor a factor of b-1, then

*p*divides n(p-1).

The claim is proved. Please note that if n(k) is a multiple of

*p*, then so is n(2k), n(3k), and so on.