2022 AIME I Problems/Problem 7
Solution 1 (sophisticated bash)
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 the numerator to be as small as possible and the denominator as big as possible. One way to do this is to make the numerator one 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 ![\[\dfrac{2 \cdot 3\cdot 6 - 7 \cdot 1 \cdot 5}{9 \cdot 8 \cdot 4}=\frac{1}{288}.\]](http://latex.artofproblemsolving.com/9/d/8/9d8a46385e2ee0ef4c08661b23781da40c9cb75d.png)
Since the maximum denominator is 
which is less than 
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 numerator 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 choices are 
(paired with either 
or 
) or 
. Neither gives us a numerator of 
so we can conclude that the minimum fraction is and thus the answer is 
.
-jgplay
