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 ~ | + | ~MRENTHUSIASM ~jgplay |
==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 17:38, 21 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
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
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.