1983 AIME Problems/Problem 6: Difference between revisions
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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. | ||
== 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 | ||