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

(Solution to 1999 AMC 8 Problem #24) |
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− | + | ==Problem 24== | |

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+ | When <math>1999^{2000}</math> is divided by <math>5</math>, the remainder is | ||

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+ | <math>\text{(A)}\ 4 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 0</math> | ||

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+ | ==Solution== | ||

+ | Note that the units digits of the powers of 9 have a pattern:<math>9^1 = ''9''</math>,<math>9^2 = 8''1''</math>,<math>9^3 = 72''9''</math>,<math>9^4 = 656''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>1</math>. |

## Revision as of 12:18, 17 June 2011

## Problem 24

When is divided by , the remainder is

## Solution

Note that the units digits of the powers of 9 have a pattern:,,,, 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 , the number ends in a . Since the exponent is even, the final digit is . Note that all natural numbers that end in have a remainder of when divided by . So, our answer is .