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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 23"

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==Problem==
 
==Problem==
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Of <math>21</math> students taking Mathematics, Physics and Chemistry, no student takes one subject only. The number of students taking Mathematics and Chemistry only, equals to four times the number taking Mathematics and Physics only. If the number of students taking Physics and Chemistry only equals to three times the number of students taking all three subjects, then the number of students taking all three subjects is
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<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 1</math>
  
 
==Solution==
 
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Let <math>x</math> be the number of students in Math and Physics and <math>y</math> be the number of those in all three. Then, <math>4x</math> is the number of students in Math and Chem, and <math>3y</math> is the number of those in Physics and Chem.
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Adding, <math>5x+4y=21</math>. We're looking for <math>y</math>, which is <math>4y\equiv21\equiv1\pmod{5}\Longrightarrow y\equiv4\pmod{5}</math>. Since <math>0\le4y<21</math>, the answer is <math>y=4\ \mathrm{(D)}</math>.
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|num-b=22|num-a=24}}
 
{{CYMO box|year=2006|l=Lyceum|num-b=22|num-a=24}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 13:18, 26 April 2008

Problem

Of $21$ students taking Mathematics, Physics and Chemistry, no student takes one subject only. The number of students taking Mathematics and Chemistry only, equals to four times the number taking Mathematics and Physics only. If the number of students taking Physics and Chemistry only equals to three times the number of students taking all three subjects, then the number of students taking all three subjects is

$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 1$

Solution

Let $x$ be the number of students in Math and Physics and $y$ be the number of those in all three. Then, $4x$ is the number of students in Math and Chem, and $3y$ is the number of those in Physics and Chem.

Adding, $5x+4y=21$. We're looking for $y$, which is $4y\equiv21\equiv1\pmod{5}\Longrightarrow y\equiv4\pmod{5}$. Since $0\le4y<21$, the answer is $y=4\ \mathrm{(D)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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