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Difference between revisions of "1999 AMC 8 Problems/Problem 24"
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==Solution==
 
==Solution==
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Note that the units digits of the powers of 9 have a pattern: <math>9^1 = {\bf 9}</math>,<math>9^2 = 8{\bf 1}</math>,<math>9^3 = 72{\bf 9}</math>,<math>9^4 = 656{\bf 1}</math>, 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 <math>9</math>, the number ends in a <math>1</math>. Since the exponent is even, the final digit is <math>1</math>. Note that all natural numbers that end in <math>1</math> have a remainder of <math>1</math> when divided by <math>5</math>. So, our answer is <math>\boxed{1}</math>.
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Note that the units digits of the powers of 9 have a pattern: <math>9^1 = {\bf 9}</math>,<math>9^2 = 8{\bf 1}</math>,<math>9^3 = 72{\bf 9}</math>,<math>9^4 = 656{\bf 1}</math>, 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 <math>9</math>, the number ends in a <math>1</math>. Since the exponent is even, the final digit is <math>1</math>. Note that all natural numbers that end in <math>1</math> have a remainder of <math>1</math> when divided by <math>5</math>. So, our answer is <math>\boxed{1, D}</math>.
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==See also==
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{{AMC8 box|year=1999|num-b=23|num-a=25}}

Revision as of 16:35, 30 July 2011

Problem 24

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 $\boxed{1, D}$.

See also

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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