Difference between revisions of "1983 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>. | Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>. | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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=== Solution 3 (cheap and quick) === | === Solution 3 (cheap and quick) === | ||
− | 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 | + | 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 is equal to <math>\boxed{35}\:\text{mod}\:49</math> |
− | == Solution 3== | + | -dragoon |
+ | |||
+ | === Solution 3=== | ||
<math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math> | <math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math> | ||
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<math>6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}</math> | <math>6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}</math> | ||
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+ | === Solution 4 last resort (bash) === | ||
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+ | Repeat the steps of taking modulo <math>49</math> after reducing the exponents over and over again until you get a residue of <math>49,</math> namely <math>35.</math> This bashing takes a lot of time but it isn’t too bad. ~peelybonehead | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/-H4n-QplQew?t=792 | ||
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+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == |
Latest revision as of 19:20, 14 January 2023
Contents
[hide]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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |