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> |
+ | |||
+ | and | ||
+ | |||
+ | <center><math> N= \{ 20 p + 16q + 12r: p,q,r \rm{ \ are \ } \rm{integers}\} </math></center> | ||
+ | |||
+ | is | ||
+ | <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
and
is
Solution
Any integer that can be created through can be created through and vice versa. Thus .