Difference between revisions of "2008 AIME II Problems/Problem 1"
(→Solution) |
m |
||
Line 34: | Line 34: | ||
\end{align*}</cmath></center> | \end{align*}</cmath></center> | ||
− | == See also == | + | == See also == |
{{AIME box|year=2008|n=II|before=First Question|num-a=2}} | {{AIME box|year=2008|n=II|before=First Question|num-a=2}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:25, 24 November 2019
Contents
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 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,

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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.