2008 AIME II Problems/Problem 1
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,
![\begin{align*} 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*}](http://latex.artofproblemsolving.com/0/b/1/0b1a194ec0e04de5792553302910e506b8737c33.png)
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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