Difference between revisions of "2008 AIME II Problems/Problem 1"
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 13:35, 19 April 2008
Problem
Let 
, where the additions and subtractions alternate in pairs. Find the remainder when 
is divided by 
.
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,
N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\ &= \underbrace{197 - 193}_4 + \underbrace{189 - 185}_4 + \cdots + \underbrace{5 - 1}_4 \\ &= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100}
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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 | ||
