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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 23"
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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>
 
<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>
  
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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>.
 
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>.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 22|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 24|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 10:49, 23 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 $M=N$.

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