2022 AIME I Problems/Problem 7
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 (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
See Also
| 2022 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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