Difference between revisions of "1983 AIME Problems/Problem 6"
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\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. | + | *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 |