Difference between revisions of "1999 AMC 8 Problems/Problem 24"

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==See Also==
 
==See Also==
 
{{AMC8 box|year=1999|num-b=23|num-a=25}}
 
{{AMC8 box|year=1999|num-b=23|num-a=25}}
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{{MAA Notice}}

Revision as of 23:34, 4 July 2013

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

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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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