Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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>35</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}}
−
 
−
* [[AIME Problems and Solutions]]
 
−
* [[American Invitational Mathematics Examination]]
 
−
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 13:40, 10 June 2008

Problem

Let $a_n$ equal $6^{n}+8^{n}$. Determine the remainder upon dividing $a_ {83}$ by $49$.

Solution

Solution 1

First, we try to find a relationship between the numbers we're provided with and $49$. We realize that $49=7^2$ and both $6$ and $8$ greater or less than $7$ by $1$.

Expressing the numbers in terms of $7$, we get $(7-1)^{83}+(7+1)^{83}$.

Applying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)$. We realize that all of the terms in this big jumble of numbers are divisible by $49$ except the final term.

After some quick division, our answer is $\boxed{035}$.

Solution 2

Since $\phi(49) = 42$ (the Euler's totient function), by Euler's Totient Theorem, $a^{42} \equiv 1 \pmod{49}$ where $\text{gcd}(a,49) = 1$. Thus $6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1}$ $\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48}$ $\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_6&oldid=26409"