Difference between revisions of "1983 AIME Problems/Problem 6"
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Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>. | Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>. | ||
− | Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of | + | Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term. |
After some quick division, our answer is <math>\boxed{035}</math>. | After some quick division, our answer is <math>\boxed{035}</math>. |
Revision as of 16:38, 1 March 2009
Problem
Let equal . Determine the remainder upon dividing by .
Contents
[hide]Solution
Solution 1
First, we try to find a relationship between the numbers we're provided with and . We realize that and both and greater or less than by .
Expressing the numbers in terms of , we get .
Applying the Binomial Theorem, half of our terms cancel out and we are left with . We realize that all of these terms are divisible by except the final term.
After some quick division, our answer is .
Solution 2
Since (the Euler's totient function), by Euler's Totient Theorem, where . Thus .
See also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |