Difference between revisions of "1983 AIME Problems/Problem 6"

 
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== Problem ==
 
== Problem ==
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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>.
  
 
== Solution ==
 
== Solution ==
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First, we try to find a relationship between the numbers we're provided with and 49. 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>.
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Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>.
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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 49 except the final term.
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After some quick division, our answer is <math>35</math>.
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----
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* [[1983 AIME Problems/Problem 5|Previous Problem]]
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* [[1983 AIME Problems/Problem 7|Next Problem]]
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* [[1983 AIME Problems|Back to Exam]]
  
 
== See also ==
 
== See also ==
* [[1983 AIME Problems]]
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 23:01, 23 July 2006

Problem

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

Solution

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 $35$.


See also