Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AIME II Problems/Problem 1"

(solution)
 
m (Style)
Line 4: Line 4:
 
== 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,
−
<cmath>\begin{align*}
+
<center><math>\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*}</cmath>
+
\end{align*}</math></center>
  
 
== See also ==
 
== See also ==

Revision as of 13:34, 19 April 2008

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*}$ (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_1&oldid=25007"