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 1

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 $\boxed{\text{(D)}\ 1}$ .

Solution 2

Write $1999$ as $2000-1$ . We are taking $(2000-1)^{2000} \mod{10}$ . Using the binomial theorem, we see that ALL terms in this expansion are divisible by $2000$ except for the very last term, which is just $(-1)^{2000}$ . This is clear because the binomial expansion is just choosing how many $2000$ s and how many $-1$ s there are for each term. Using this, we can take the entire polynomial $\mod{10}$ , which leaves just $(-1)^{2000}=\boxed{\text{(D)}\ 1}$ .

Solution 3

As $1999 \equiv -1 \pmod{5}$ , we have $1999^{2000} \equiv (-1)^{2000} \equiv 1 \pmod{5}$ . Thus, the 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

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1999 AMC 8 Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also