Difference between revisions of "2022 AIME I Problems/Problem 7"
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Together, we conclude that the minimum possible positive value of <math>\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}</math> is <math>\frac{1}{288}.</math> Therefore, the answer is <math>1+288=\boxed{289}.</math> | Together, we conclude that the minimum possible positive value of <math>\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}</math> is <math>\frac{1}{288}.</math> Therefore, the answer is <math>1+288=\boxed{289}.</math> | ||
− | ~MRENTHUSIASM ~jgplay | + | ~MRENTHUSIASM ~jgplay ~PureSwag |
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=6|num-a=8}} | {{AIME box|year=2022|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:57, 7 March 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
To minimize a positive fraction, we minimize its numerator and maximize its denominator. We must have 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 ~PureSwag
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.