AoPSWiki

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

Personal tools

Search

Toolbox

University of South Carolina High School Math Contest/1993 Exam/Problem 23

From AoPSWiki

Problem

The relation between the sets

$M = \{ 12 m + 8 n + 4 l: m,n,l \rm{ \ are \ } \rm{integers}\}$

and

$N= \{ 20 p + 16q + 12r: p,q,r \rm{ \ are \ } \rm{integers}\}$

is

$\mathrm{(A) \ } M\subset N \qquad \mathrm{(B) \ } N\subset M \qquad \mathrm{(C) \ } M\cup N = \{0\} \qquad \mathrm{(D) \ }602...$

Solution

Every element of $M$ is a multiple of 4, and any multiple of 4 is an element of $M$ (set $m = n = 0$ and choose $l$ as needed). Every element of $N$ is also a multiple of 4. If $k$ is an integer, $4k = 20\cdot 0 + 16\cdot k + 12 \cdot (-k)$ so that every multiple of 4 is an element of $N$ . Since $M$ and $N$ have the same elements, they are the same set: $M = N \Longrightarrow \mathrm{(E)}$ .

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_23"

Category: Introductory Number Theory Problems

Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!

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