Problem:
Prove that there exists a positive integer n<106 such that 5n has six consecutive zeros in its decimal representation.
Solution:
One answer is n=20+219=524308.
First, observe that
​5n≡520(mod520)5n≡520(mod220)​
the former being immediate and the latter since φ(220)=219. Hence 5n≡520(mod1020). Moreover, we have
520=2201​⋅1020<100021​⋅1020=10−6⋅1020
Thus the last 20 digits of 5n will begin with six zeros. This completes the proof.
The problems on this page are the property of the MAA's American Mathematics Competitions