1983 AIME Problems/Problem 6
Contents
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 (cheap and quick)
As the value of is obviously we look for a pattern with others. With a bit of digging, we discover that where and are odd is equal to
-dragoon
Solution 3
Becuase , we only consider
Solution 4 last resort (bash)
Repeat the steps of taking modulo after reducing the exponents over and over again until you get a residue of namely This bashing takes a lot of time but it isn’t too bad. ~peelybonehead
Video Solution by OmegaLearn
https://youtu.be/-H4n-QplQew?t=792
~ pi_is_3.14
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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