Difference between revisions of "2008 AIME II Problems/Problem 1"
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Dividing <math>10100</math> by <math>1000</math> yields a remainder of <math>\boxed{100}</math>. | Dividing <math>10100</math> by <math>1000</math> yields a remainder of <math>\boxed{100}</math>. | ||
| − | == Solution == | + | == Solution 2 == |
Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms, | Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms, | ||
<center><cmath>\begin{align*} | <center><cmath>\begin{align*} | ||
Revision as of 12:21, 6 June 2019
Contents
Problem
Let 
, where the additions and subtractions alternate in pairs. Find the remainder when 
is divided by 
.
Solution
Rewriting this sequence with more terms, we have
Factoring this expression yields
Next, we get
Then,
Dividing 
by yields a remainder of 
.
Solution 2
Since we want the remainder when is divided by , we may ignore the 
term. Then, applying the difference of squares factorization to consecutive terms,
See also
| 2008 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
