Prove the following statement:
In any set of 26 integers chosen from the set of (1,2,3, ...50) there must be at least a pair of numbers such that one of them divides the other.
(In reply to @ Steve - toward reasonable proof
by Ady TZIDON)
(21,48) was in fact a typo. I have corrected it to (21,42) in the original post. Thanks.
After the typo is corrected, my proof (using the pigeonhole principle) is correct. It is not possible to pick 26 numbers without picking more than one from at least one of the sets, and that means than any set of 26 has at least one number that divides another.
The set (1,2,4,8,16,32) is correct and intentional. The point is that I cannot pick two of those without having one divide another. I am not claiming (and I do not need to claim) that I can get a valid set of 25 by taking the 10 that are not in any of my 15 subsets and then picking any one from each set. Obviously, 1 and 2 are not present in any valid set of 25.