Difference between revisions of "1983 AIME Problems/Problem 6"
Sevenoptimus (talk | contribs) (Cleaned up the solutions) |
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*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. | ||
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+ | == Solution 3== | ||
+ | <math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math> | ||
+ | |||
+ | Becuase <math>7|(6+8)</math>, we only consider <math>(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82}) \text{mod} 7</math> | ||
== See Also == | == See Also == |
Revision as of 12:52, 9 March 2021
Problem
Let . Determine the remainder upon dividing by .
Solution
Solution 1
Firstly, we try to find a relationship between the numbers we're provided with and . We notice that , and both and are greater or less than by .
Thus, 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 (see Euler's totient function), Euler's Totient Theorem tells us that where . Thus .
- Alternatively, we could have noted that . This way, we have , and can finish the same way.
Solution 3
Becuase , we only consider
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 |