Difference between revisions of "1983 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
+ | 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 == | ||
+ | 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>. | ||
+ | |||
+ | Expressing the numbers in terms of <math>7</math>, we get <math>(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 the terms in this big jumble of numbers are divisible by 49 except the final term. | ||
+ | |||
+ | After some quick division, our answer is <math>35</math>. | ||
+ | |||
+ | ---- | ||
+ | |||
+ | * [[1983 AIME Problems/Problem 5|Previous Problem]] | ||
+ | * [[1983 AIME Problems/Problem 7|Next Problem]] | ||
+ | * [[1983 AIME Problems|Back to Exam]] | ||
== See also == | == See also == | ||
− | * [[ | + | * [[AIME Problems and Solutions]] |
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 23:01, 23 July 2006
Problem
Let equal . Determine the remainder upon dividing by .
Solution
First, we try to find a relationship between the numbers we're provided with and 49. We realize that and both and greater or less than 7 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 49 except the final term.
After some quick division, our answer is .