Difference between revisions of "1983 AIME Problems/Problem 6"
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As the value of <math>a</math> is obviously <math>6^83+8^83</math> we look for a pattern with others. With a bit of digging, we discover that <math>6^n+6^m</math> where <math>m</math> and <math>n</math> are odd, when divided by <math>49</math> it has a remainder of <math>35</math> (I don’t know how to do modular arithmetic in <math>LaTeX</math> | As the value of <math>a</math> is obviously <math>6^83+8^83</math> we look for a pattern with others. With a bit of digging, we discover that <math>6^n+6^m</math> where <math>m</math> and <math>n</math> are odd, when divided by <math>49</math> it has a remainder of <math>35</math> (I don’t know how to do modular arithmetic in <math>LaTeX</math> | ||
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== Solution 3== | == Solution 3== |
Revision as of 16:47, 29 January 2022
Problem
Let . Determine the remainder upon dividing by .
Contents
[hide]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, when divided by it has a remainder of (I don’t know how to do modular arithmetic in
-dragoon
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 |