2008 AIME II Problems/Problem 1
Problem
Let , where the additions and subtractions alternate in pairs. Find the remainder when is divided by .
Solution 1
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,
Solution 3
By observation, we realize that the sequence alternates every 4 terms. Simplifying, we get , turning into a arithmetic sequence with 25 terms, them being , as the series alternates every 4 terms.
Applying the sum of arithmetic sequence formula, we get
So the answer would be .
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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 |
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