Difference between revisions of "1966 IMO Problems/Problem 1"
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PLEASE PROVIDE A 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 = | + | 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}} |
Revision as of 22:46, 28 April 2014
In a mathematical contest, three problems, ,
, and
were posed. Among the participants there were
students who solved at least one problem each. Of all the contestants who did not solve problem
, the number who solved
was twice the number who solved
. The number of students who solved only problem
was one more than the number of students who solved
and at least one other problem. Of all students who solved just one problem, half did not solve problem
. How many students solved only problem
?
Solution
Let us draw a Venn Diagram.
PLEASE PROVIDE A DIAGRAM.
Let be the number of students solving both B and C. Then for some positive integer
,
students solved B only, and
students solved C only. Let
be the number of students solving A; then
is the number of students solving A only. We have by given
and
Substituting for y into the first equation gives
Thus, because
and
are positive integers with
, we have
and
. (Note that
and
does not work.) Hence, the number of students solving B only is
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 |