1999 AMC 8 Problems/Problem 24

Problem

When $1999^{2000}$ is divided by $5$ , the remainder is

$\text{(A)}\ 4 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 0$

Solution

Note that the units digits of the powers of 9 have a pattern: $9^1 = {\bf 9}$ , $9^2 = 8{\bf 1}$ , $9^3 = 72{\bf 9}$ , $9^4 = 656{\bf 1}$ , and so on. Since all natural numbers with the same last digit have the same remainder when divided by 5, the entire number doesn't matter, just the last digit. For even powers of $9$ , the number ends in a $1$ . Since the exponent is even, the final digit is $1$ . Note that all natural numbers that end in $1$ have a remainder of $1$ when divided by $5$ . So, our answer is $\boxed{\text{(D)}\ 1}$ .

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1999 AMC 8 Problems/Problem 24

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