Difference between revisions of "1983 AIME Problems/Problem 6"
Sevenoptimus (talk | contribs) m (Fixed problem statement) |
Sevenoptimus (talk | contribs) (Cleaned up the solutions) |
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | + | Firstly, we try to find a relationship between the numbers we're provided with and <math>49</math>. We notice that <math>49=7^2</math>, and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>. | |
− | + | Thus, expressing the numbers in terms of <math>7</math>, we get <math>a_{83} = (7-1)^{83}+(7+1)^{83}</math>. | |
Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term. | Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term. | ||
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=== Solution 2 === | === Solution 2 === | ||
− | Since <math>\phi(49) = 42</math> ( | + | Since <math>\phi(49) = 42</math> (see [[Euler's totient function]]), [[Euler's Totient Theorem]] tells us that <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math> |
<math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math> | <math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math> | ||
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>. | \equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>. |
Revision as of 18:24, 15 February 2019
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.
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 |