# Pigeon hole principle sum of integers?

You randomly select $$k$$ integers between $$1$$ and $$100$$, inclusive. What is the smallest $$k$$ that guarantees that at least one pair of the selected integers will sum to 101?

I have to use the pigeon hole principle to solve the problem. My attempt at the solution is to choose $$k=51$$ integers from $$1$$ to $$51$$. The pair $$50$$ and $$51$$ add up to $$101$$. My question is why would I start at 1? Could I just start at $$k = 50$$ and stop at $$51$$ and I have my sum that adds up to $$101$$. Other pairs if I started at $$49$$ and stopped at $$52$$ I could still choose $$49$$ and $$52$$ and that would still add up to $$101$$. $$k$$ would be $$4$$ integers instead of $$2$$. From my reasoning the smallest amount of $$k$$ integers I would have to pick would be two? I still don't understand how I would apply the pigeon hole principle here if I can start from fifty. Any help would be appreciated.

First off, your method of choosing some numbers is incorrect. $$50$$ and $$51$$ do add to $$101$$, but if $$k=2$$, then the question would say that whatever two numbers we pick, they would sum to $$101$$. But if we say, chose $$1$$ and $$2$$, these clearly don't add to $$101$$. What we want is no matter which $$k$$ numbers we pick, we can find two that sum to $$101$$.
What I mean by this, is instead of $$100$$ numbers from $$1$$ to $$100$$, we now have $$50$$ pairs $$\{(1,100), (2,99), (3,98), ..., (50,51) \}$$ And now, it should become apparent that in order to choose no pairs that sum to $$101$$, we must choose a number from each pair at most once. Thus, when we choose numbers from $$51$$ pairs, by pidgeonhole principle, we will have chosen both numbers from some pair, and thus have two numbers adding to $$101$$. Hope that helps!