2022 AIME I Problems/Problem 7
Contents
Problem
Let 
be distinct integers from 
to 
The minimum possible positive value of ![\[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\]](http://latex.artofproblemsolving.com/c/6/5/c653d108c03120ef617df93994006dd9a956c44c.png)
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
Solution 2
Since we are trying to minimize ![\[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i},\]](http://latex.artofproblemsolving.com/4/f/6/4f6945117120779cbfa0bd24ea5872edb5d8e3a6.png)
we want to minimize its numerator and maximize its denominator. One way to do this is to make the numerator 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 ![\[\dfrac{2 \cdot 3\cdot 6 - 1 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 9}=\frac{1}{288}.\]](http://latex.artofproblemsolving.com/5/2/7/527a14dad3e2d264fd377f32ca03dba13ae8adf7.png)
Since the maximum denominator is 
we conclude that 
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 denominator 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 choice is one of
(paired with 
)
(paired with 
)
(paired with either 
or 
)
(paired with either 
or 
)
(paired with 
)
(paired with 
)
(paired with 
)
(paired with 
)None of them gives us a numerator of 
so the minimum fraction is and thus the answer is
~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 | ||
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