Difference between revisions of "1983 AIME Problems/Problem 6"
| Line 1: | Line 1: | ||
== 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 
.
