Difference between revisions of "1993 AIME Problems/Problem 7"
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'''Note''' that the <math>1000</math> in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers <math>x_1,x_2,x_3,x_4,x_5,x_6</math> whether they may be <math>1,2,3,4,5,6</math> or <math>999,5,3,998,997,891</math>. | '''Note''' that the <math>1000</math> in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers <math>x_1,x_2,x_3,x_4,x_5,x_6</math> whether they may be <math>1,2,3,4,5,6</math> or <math>999,5,3,998,997,891</math>. | ||
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== See also == | == See also == |
Revision as of 20:23, 17 April 2015
Problem
Three numbers, ,
,
, are drawn randomly and without replacement from the set
. Three other numbers,
,
,
, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let
be the probability that, after a suitable rotation, a brick of dimensions
can be enclosed in a box of dimensions
, with the sides of the brick parallel to the sides of the box. If
is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Solution 1
Call the six numbers selected . Clearly,
must be a dimension of the box, and
must be a dimension of the brick.
- If
is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us
possibilities.
- If
is not a dimension of the box but
is, then both remaining dimensions will work as a dimension of the box. That gives us
possibilities.
- If
is a dimension of the box but
aren’t, there are no possibilities (same for
).
The total number of arrangements is ; therefore,
, and the answer is
.
Note that the in the problem, is not used, and is cleverly bypassed in the solution, because we can call our six numbers
whether they may be
or
.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.