Problem:
Find the remainder when
9⋅99⋅999⋯9999′ s99…9
is divided by 1000 .
Solution:
Let N be the product in the problem. Then
N≡9⋅99(−1)997≡−891≡109(mod1000)
Thus the desired remainder is 109.
The problems on this page are the property of the MAA's American Mathematics Competitions