Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 23"
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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]] | ||
− | [[Category: | + | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 14:06, 19 August 2006
Problem
The relation between the sets
![$M = \{ 12 m + 8 n + 4 l: m,n,l \rm{ \ are \ } \rm{integers}\}$](http://latex.artofproblemsolving.com/8/5/d/85df64d0e050415fe93cde4d432162aea5546227.png)
and
![$N= \{ 20 p + 16q + 12r: p,q,r \rm{ \ are \ } \rm{integers}\}$](http://latex.artofproblemsolving.com/a/1/3/a137cf2604fac5c5ae08b650f9bc1db1f29575e2.png)
is
![$\mathrm{(A) \ } M\subset N \qquad \mathrm{(B) \ } N\subset M \qquad \mathrm{(C) \ } M\cup N = \{0\} \qquad \mathrm{(D) \ }60244 \rm{ \ is \ } \rm{in \ } M \rm{ \ but \ } \rm{not \ } \rm{in \ } N \qquad \mathrm{(E) \ } M=N$](http://latex.artofproblemsolving.com/a/6/6/a66382490281eeb37a27763b42602a93e82b8126.png)
Solution
Every element of is a multiple of 4, and any multiple of 4 is an element of
(set
and choose
as needed). Every element of
is also a multiple of 4. If
is an integer,
so that every multiple of 4 is an element of
. Since
and
have the same elements, they are the same set:
.