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>. | ||
== Solution == | == Solution == | ||
+ | === 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. | ||
+ | |||
+ | After some quick division, our answer is <math>\boxed{035}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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> | ||
+ | \equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>. | ||
+ | |||
+ | *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 (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 is equal to <math>\boxed{35}\:\text{mod}\:49</math> | ||
+ | |||
+ | -dragoon | ||
+ | |||
+ | === 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} \pmod{7}</math> | ||
+ | |||
+ | <math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \equiv (-1)^{82} - (-1)^{81}+ \ldots - (-1)^1 + 1 = 83 \equiv 6 \pmod{7}</math> | ||
+ | |||
+ | <math>6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}</math> | ||
+ | |||
+ | |||
+ | |||
+ | === Solution 4 last resort (bash) === | ||
+ | |||
+ | 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 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/-H4n-QplQew?t=792 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == See Also == | ||
+ | {{AIME box|year=1983|num-b=5|num-a=7}} | ||
− | + | [[Category:Intermediate Number Theory Problems]] | |
− |
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 |