Difference between revisions of "2008 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms, | Since we want the remainder when <math>N</math> is divided by <math>1000</math>, we may ignore the <math>100^2</math> term. Then, applying the [[difference of squares]] factorization to consecutive terms, | ||
| − | <center>< | + | <center><cmath>\begin{align*} |
N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\ | 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 \\ | &= \underbrace{197 - 193}_4 + \underbrace{189 - 185}_4 + \cdots + \underbrace{5 - 1}_4 \\ | ||
&= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100} | &= 4 \cdot \left(\frac{197-5}{8}+1\right) = \boxed{100} | ||
| − | \end{align*}</ | + | \end{align*}</cmath></center> |
== See also == | == See also == | ||
Revision as of 02:06, 3 March 2015
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,
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. 
