Difference between revisions of "1966 IMO Problems/Problem 1"

m
Line 2: Line 2:
  
 
==Solution==
 
==Solution==
{{solution}}
+
Let us draw a Venn Diagram.
 +
 
 +
PLEASE PROVIDE A DIAGRAM.
 +
 
 +
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 = 24</cmath> and <cmath>y = 3x - 2a.</cmath> Substituting for y into the first equation gives <cmath>9x - 5a = 25.</cmath> Thus, because <math>x</math> and <math>a</math> are positive integers with <math>x-a \ge 0</math>, we have <math>x = 5</math> and <math>a = 4</math>. (Note that <math>x = 10</math> and <math>a = 13</math> does not work.) Hence, the number of students solving B only is <math>2x - a = 10 - 4 = \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}}

Revision as of 23:45, 28 April 2014

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.

PLEASE PROVIDE A DIAGRAM.

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 \[2y - 1 + 3x - a = 24\] and \[y = 3x - 2a.\] Substituting for y into the first equation gives \[9x - 5a = 25.\] Thus, because $x$ and $a$ are positive integers with $x-a \ge 0$, we have $x = 5$ and $a = 4$. (Note that $x = 10$ and $a = 13$ does not work.) Hence, the number of students solving B only is $2x - a = 10 - 4 = \boxed{6}.$

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
Invalid username
Login to AoPS