AoPSWiki

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

Personal tools

Search

Toolbox

1991 AJHSME Problems/Problem 4

From AoPSWiki

Problem

If $991+993+995+997+999=5000-N$ , then $N=$

$\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$

Solution

$\begin{align*}991+993+995+997+999=5000-N &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\&\Right...$

See Also

1991 AJHSME (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1991_AJHSME_Problems/Problem_4"

Category: Introductory Algebra Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us