Difference between revisions of "1983 AIME Problems/Problem 6"
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=== Solution 2 === | === Solution 2 === | ||
− | Since <math>\phi(49) = 42</math> (the [[Euler's totient function]]), by [[Euler's Totient Theorem]], <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math> <math> | + | Since <math>\phi(49) = 42</math> (the [[Euler's totient function]]), by [[Euler's Totient Theorem]], <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math> |
− | \equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math> | + | <math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math> |
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>. | \equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>. | ||
+ | |||
+ | *Alternatively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{n}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}</math>, and can finish the same way. | ||
== See Also == | == See Also == |
Revision as of 17:51, 16 June 2014
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 are 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 .
- Alternatively, we could have noted that . This way, we have , and can finish the same way.
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 |