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><math>\begin{align*}
+
<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*}</math></center>
+
\end{align*}</cmath></center>
  
 
== See also ==
 
== See also ==

Revision as of 03:06, 3 March 2015

Problem

Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$, where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$.

Solution

Since we want the remainder when $N$ is divided by $1000$, we may ignore the $100^2$ 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*}

See also

2008 AIME II (ProblemsAnswer KeyResources)
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

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