Difference between revisions of "1983 AIME Problems/Problem 6"
m (box) |
(alt. solution) |
||
Line 2: | Line 2: | ||
Let <math>a_n</math> equal <math>6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>. | Let <math>a_n</math> equal <math>6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>. | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | First, we try to find a relationship between the numbers we're provided with and <math>49</math>. We realize that <math>49=7^2</math> and both <math>6</math> and <math>8</math> greater or less than 7 by <math>1</math>. | + | === Solution 1 === |
+ | First, we try to find a relationship between the numbers we're provided with and <math>49</math>. We realize that <math>49=7^2</math> and both <math>6</math> and <math>8</math> greater or less than <math>7</math> by <math>1</math>. | ||
Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>. | Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>. | ||
Line 9: | Line 11: | ||
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 the terms in this big jumble of numbers 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 the terms in this big jumble of numbers are divisible by <math>49</math> except the final term. | ||
− | After some quick division, our answer is <math> | + | After some quick division, our answer is <math>\boxed{035}</math>. |
+ | |||
+ | === Solution 2 === | ||
+ | Since <math>\phi(49) = 42</math> (the [[Euler's totient function]]), by [[Euler's Totient Theorem]], <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>. | ||
== See also == | == See also == | ||
{{AIME box|year=1983|num-b=5|num-a=7}} | {{AIME box|year=1983|num-b=5|num-a=7}} | ||
− | |||
− | |||
− | |||
− | |||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |
Revision as of 13:40, 10 June 2008
Problem
Let equal . Determine the remainder upon dividing by .
Contents
[hide]Solution
Solution 1
First, we try to find a relationship between the numbers we're provided with and . We realize that and both and greater or less than by .
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 the terms in this big jumble of numbers are divisible by except the final term.
After some quick division, our answer is .
Solution 2
Since (the Euler's totient function), by Euler's Totient Theorem, where . Thus .
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 |