At least how many numbers should be selected from the set {1, 5, 9, 13, ...125} to be assured that two of the numbers selected have a sum of 146? I know the answer is 20 (says the answer key), but I'm not quite sure how it got it.   
I also know that in the sequence, we can pair $21+125$, $25+121$, $29+117$... and so on to get a sum of $146$. From 21 to 125, there are 27 terms, which means there are 13 pairs in the set that make a sum of 146. From here, I don't know how to go on any further.
 A: The numbers that can't be part of any pair summing to 146 are 1, 5, 9, 13, 17 and 73 -- there are 6 of these and in the worst case all of them will be selected. Thus, if you select 20 numbers, at least 14 of them will be chosen among the 2×13 numbers that form useful pairs, and by the pigeonhole principle at least one of the pairs will have both its members selected.
On the other hand, you can select 19 numbers such that no two sum to 146, for example all the ones from 1 up to 73.
A: This is a textbook case of the Pigeonhole principle. You already correctly identified that are 13 containers, each containing two numbers, such that selecting both numbers will yield a sum of 146: $\{21, 125\}, \{25, 121\},\dots, \{69, 77\}$. We also have 6 containers, each containing one number, that will never yield a sum of 146: $\{1\}, \{5\},\dots,\{17\}, \{73\}$.
So, if we select 19 numbers from these containers, we could get one from each of the first 13 and all 6 from the others, not yielding 146.
If instead we select 20 we have guaranteed that we will select 2 from one of the first 13, yielding 146.
