1999 AMC 8 Problem 24

Attempt Problem 24 of the 1999 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

When 199920001999^{2000} is divided by 5,5, the remainder is
44
33
22
11
00
Answer: D
Concepts:modular arithmeticunits digit
Small hint:
Reduce 19991999 modulo 55
Big hint:
Even powers of −1-1 modulo 55 have the same remainder
Solution:
Note that to find the remainder when divided by 5,5, we only care about the units digit.
This means we only have to observe how the powers of the units digit work, namely the powers of 9.9.
Then, looking at powers of 9,9, we see that the units digit alternates between 99 and 1:1: 9,81,729,⋯ . 9, 81, 729, \cdots. This means that 199920001999^{2000} ends in a 11 since the power is even.
The remainder when divided by 55 is then 1,1, since it is 11 more than a multiple of 10.10.
Thus, D is the correct answer.
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