Difference between revisions of "2008 AIME II Problems/Problem 1"
(→Solution 2) |
m (→Solution) |
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Line 20: | Line 20: | ||
Then, | Then, | ||
<center><cmath>\begin{align*} | <center><cmath>\begin{align*} | ||
− | N &= 2(\frac{(100)(101)}{2} \mbox{, and simplifying, we get}\ | + | N &= 2\left(\frac{(100)(101)}{2}\right) \mbox{, and simplifying, we get}\ |
N &= (100)(101) \mbox{, so}\ | N &= (100)(101) \mbox{, so}\ | ||
N &= 10100 \mbox{.} | N &= 10100 \mbox{.} |
Revision as of 21:10, 26 June 2017
Contents
[hide]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
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.