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> | ||
== Solution == | == Solution == | ||
− | + | 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/Problem 22|Previous Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 24|Next Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
− | + | [[Category:Introductory Number Theory Problems]] | |
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Latest revision as of 14:06, 19 August 2006
Problem
The relation between the sets
and
is
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: .