Difference between revisions of "1983 AIME Problems/Problem 6"
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== 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>. | + | 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>. |
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>. | ||
| − | 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. | + | 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>35</math>. | ||
| − | -- | + | == See also == |
| + | {{AIME box|year=1983|num-b=5|num-a=7}} | ||
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* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
Revision as of 17:37, 21 March 2007
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 . 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 except the final term.
After some quick division, our answer is 
.
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 | ||
