Problem:
A teacher was leading a class of four perfectly logical students. The teacher chose a set of four integers and gave a different number in to each student. Then the teacher announced to the class that the numbers in were four consecutive two-digit positive integers, that some number in was divisible by , and a different number in was divisible by . The teacher then asked if any of the students could deduce what is, but in unison, all of the students replied no. However, upon hearing that all four students replied no, each student was able to determine the elements of . Find the sum of all possible values of the greatest element of .
Solution:
Let's do casework based off the distance between the multiples of 6 and 7.
Notice that one person will receive a multiple of 6, one a multiple of 7, and 2 neither.
Suppose that they are distance 2, so one number is between them.
In that case, the person that is between the multiple of 6 and 7 won't be able to deduce it, because the last number could either be on the left side of S or the right side of S, but the person that is not between will be able to directly determine it.
Now, suppose they are distance 3, so two numbers are between them.
In this case, the people who have a multiple of 6, or the two that have neither will be able to instantly decipher the set S, but the person with the multiple of 7 won't be able to, but either way this case doesn't work.
The last and only valid case is when the multiples of 6 and 7 are next to each other. In this case, it could have the two non-multiples greater/smaller than the multiples, and the other case is when one is less than a multiple, and the other is greater than the other multiple.
In the case where the two numbers are greater/smaller, notice how the "endmost" member that has a number that's not a multiple of 6/7 will immediately be able to figure out the list, so this case doesn't work, and the only set of valid cases is when the two multiples are adjacent, with one number on each side.
Enumerating these yields
, so our answer is
The problems on this page are the property of the MAA's American Mathematics Competitions