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

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== Solution ==
 
== Solution ==
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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>.
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Every [[element]] of <math>M</math> is a [[multiple]] of 4, and any multiple of 4 is an element of <math>M</math> (set <math>m = n = 0</math> and choose <math>l</math> as needed).  Every element of <math>N</math> is also a multiple of 4.  If <math>k</math> is an integer, <math>4k = 20\cdot 0 + 16\cdot k + 12 \cdot (-k)</math> so that every multiple of 4 is an element of <math>N</math>.  Since <math>M</math> and <math>N</math> have the same elements, they are the same [[set]]: <math>M = N \Longrightarrow \mathrm{(E)}</math>.
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
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[[Category:Intermediate Number Theory Problems]]
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 14:06, 19 August 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

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: $M = N \Longrightarrow \mathrm{(E)}$.

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