Difference between revisions of "2008 AIME II Problems/Problem 1"
I like pie (talk | contribs) m (→See also) |
|||
| Line 14: | Line 14: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 22:33, 4 July 2013
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
