Difference between revisions of "2022 AIME I Problems/Problem 7"
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Revision as of 21:15, 17 February 2022
Problem
Let be distinct integers from
to
The minimum possible positive value of
can be written as
where
and
are relatively prime positive integers. Find
Solution 1 (sophisticated bash)
Since we are trying to minimize we want the numerator to be as small as possible and the denominator as big as possible. One way to do this is to make the numerator one and the denominator as large as possible. This means that
has to be a different parity than
Using this, and reserving
and
for the denominator, we notice that
Since the maximum denominator is
which is less than
will be less than any other fraction we can come up with with a numerator greater than
This means that all we need to check is fractions with numerator
and numerator greater than
. The only alternatives we need to consider are
and
in the denominator. The parity restriction allows us to focus on numerators where either
are all odd or
are all odd, so our choices are
(paired with either
or
) or
. Neither gives us a numerator of
so we can conclude that the minimum fraction is
and thus the answer is
.
-jgplay
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.