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Difference between revisions of "1966 IMO Problems/Problem 1"
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In a mathematical contest, three problems, <math>A</math>, <math>B</math>, and <math>C</math> were posed. Among the participants ther were <math>25</math> students who solved at least one problem each. Of all the contestants who did not solve problem <math>A</math>, the number who solved <math>B</math> was twice the number who solved <math>C</math>. The number of students who solved only problem <math>A</math> was one more than the number of students who solved <math>A</math> and at least one other problem. Of all students who solved just one problem, half did not solve problem <math>A</math>. How many students solved only problem <math>B</math>?
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==Problem==
  
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==See Also==
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In a mathematical contest, three problems, <math>A</math>, <math>B</math>, and <math>C</math> were posed. Among the participants there were <math>25</math> students who solved at least one problem each. Of all the contestants who did not solve problem <math>A</math>, the number who solved <math>B</math> was twice the number who solved <math>C</math>. The number of students who solved only problem <math>A</math> was one more than the number of students who solved <math>A</math> and at least one other problem. Of all students who solved just one problem, half did not solve problem <math>A</math>. How many students solved only problem <math>B</math>?
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{{solution}}
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==Solution==
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Let us draw a Venn Diagram.
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[[File:IMO1966.png]]
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Let <math>a</math> be the number of students solving both B and C. Then for some positive integer <math>x</math>, <math>2x - a</math> students solved B only, and <math>x - a</math> students solved C only. Let <math>2y - 1</math> be the number of students solving A; then <math>y</math> is the number of students solving A only. We have by given <cmath>2y - 1 + 3x - a = 25</cmath> and <cmath>y = 3x - 2a.</cmath> Substituting for y into the first equation gives <cmath>9x - 5a = 26.</cmath> Thus, because <math>x</math> and <math>a</math> are positive integers with <math>x-a \ge 0</math>, we have <math>x = 4</math> and <math>a = 2</math>. (Note that <math>x = 9</math> and <math>a = 11</math> does not work.) Hence, the number of students solving B only is <math>2x - a = 8 - 2 = \boxed{6}.</math>
  
 
==See Also==
 
==See Also==
 
{{IMO box|year=1966|before=First Question|num-a=2}}
 
{{IMO box|year=1966|before=First Question|num-a=2}}

Latest revision as of 19:18, 10 November 2024

Problem

In a mathematical contest, three problems, $A$, $B$, and $C$ were posed. Among the participants there were $25$ students who solved at least one problem each. Of all the contestants who did not solve problem $A$, the number who solved $B$ was twice the number who solved $C$. The number of students who solved only problem $A$ was one more than the number of students who solved $A$ and at least one other problem. Of all students who solved just one problem, half did not solve problem $A$. How many students solved only problem $B$?


Solution

Let us draw a Venn Diagram.

IMO1966.png

Let $a$ be the number of students solving both B and C. Then for some positive integer $x$, $2x - a$ students solved B only, and $x - a$ students solved C only. Let $2y - 1$ be the number of students solving A; then $y$ is the number of students solving A only. We have by given \[2y - 1 + 3x - a = 25\] and \[y = 3x - 2a.\] Substituting for y into the first equation gives \[9x - 5a = 26.\] Thus, because $x$ and $a$ are positive integers with $x-a \ge 0$, we have $x = 4$ and $a = 2$. (Note that $x = 9$ and $a = 11$ does not work.) Hence, the number of students solving B only is $2x - a = 8 - 2 = \boxed{6}.$

See Also

1966 IMO (Problems) • Resources
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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