Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 23"

 
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== Problem ==
 
== Problem ==
 +
The relation between the sets
  
<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
+
<center><math> M = \{ 12 m + 8 n + 4 l: m,n,l \rm{ \ are \ } \rm{integers}\} </math></center>
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 +
and
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<center><math> N= \{ 20 p + 16q + 12r: p,q,r \rm{ \ are \ } \rm{integers}\} </math></center>
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is
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<center><math> \mathrm{(A) \ } M\subset N \qquad \mathrm{(B) \ } N\subset M \qquad \mathrm{(C) \ } M\cup N = \{0\} \qquad \mathrm{(D) \ }60244 \rm{ \ is \ } \rm{in \ } M \rm{ \ but \ } \rm{not \ } \rm{in \ } N \qquad \mathrm{(E) \ } M=N </math></center>
  
 
== Solution ==
 
== Solution ==
 +
Any integer that can be created through <math>M</math> can be created through <math>N</math> and vice versa. Thus <math>M=N</math>.
  
 
== See also ==
 
== See also ==
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam]]

Revision as of 19:46, 22 July 2006

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

$\mathrm{(A) \ } M\subset N \qquad \mathrm{(B) \ } N\subset M \qquad \mathrm{(C) \ } M\cup N = \{0\} \qquad \mathrm{(D) \ }60244 \rm{ \ is \ } \rm{in \ } M \rm{ \ but \ } \rm{not \ } \rm{in \ } N \qquad \mathrm{(E) \ } M=N$

Solution

Any integer that can be created through $M$ can be created through $N$ and vice versa. Thus $M=N$.

See also