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 (Optimization)
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
Solution 2 (Bash)
Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.
To bash efficiently, we can start out with 
as our denominator. This, however, leaves us with the numbers 
and 
left. The smallest we can make out of this is 
. When simplified, it gives us 
, which gives a small answer of 
. Obviously there are larger answers than this.
After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple 
and 
.
This gives us a numerator of 
, which is by far the smallest yet. With the remaining numbers 
and 
, we get 
.
Finally, we add up our numerator and denominator: The answer is 
.
Solution 3 (Educated Trial and Error)
To minimize the numerator, we must have 
. Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value 
cannot result in a difference of , and the next smallest product, 
cannot either, but 
can if 
. Thus, the denominator must be 
, and the smallest fraction possible is 
, making the answer 
.
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. 
