Difference between revisions of "2022 AIME I Problems/Problem 7"
MRENTHUSIASM (talk | contribs) m (→Solution 2) |
MRENTHUSIASM (talk | contribs) (→Solution 2: Tried my very best making this solution more rigorous, but I realized that there are many more cases for the denominator: 5*7*9, 5*8*9, 6*7*9, 6*8*9, 6*7*8, 7*8*9. So, I decide to remove this sol and combine credits in Sol 1.) |
||
Line 13: | Line 13: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
==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 18:33, 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 1
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
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.