Difference between revisions of "2022 AIME I Problems/Problem 7"
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~MRENTHUSIASM ~jgplay | ~MRENTHUSIASM ~jgplay | ||
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Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator. | Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator. | ||
Revision as of 00:05, 1 January 2023
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 (Optimization)
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that
If we minimize the numerator, then Note that
so
It follows that
and
are consecutive composites with prime factors no other than
and
The smallest values for
and
are
and
respectively. So, we have
and
from which
If we do not minimize the numerator, then Note that
Together, we conclude that the minimum possible positive value of is
Therefore, the answer is
~MRENTHUSIASM ~jgplay
Solution 2 (Experimentation)
Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.
To bash efficiently, we can start out with as our denominator. This, however, leaves us with the numbers
and
left. The smallest we can make out of this is
. When simplified, it gives us
, which gives a small answer of
. Obviously there are larger answers than this.
After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple and
.
This gives us a numerator of , which is by far the smallest yet. With the remaining numbers
and
, we get
.
Finally, we add up our numerator and denominator: The answer is .
See Also
2022 AIME I (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.