2022 AIME I Problems/Problem 7
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 (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
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 .
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