# Show that if seven integers are selected from the ﬁrst 10 positive integers

a) Show that if seven integers are selected from the ﬁrst 10 positive integers, there must be at least two pairs of these integers with the sum 11.

b) Is the conclusion in part (a) true if six integers are selected rather than seven?

I don't know how should I show that.

I know that in the worst situation we choose 1,2,...,7 that we have 7 + 4 = 11 and 5 + 6 = 11 but I don't know what should be the answer of this question

Partition numbers from $1$ to $10$ into the following sets: $\{1,10\}$, $\{2,9\}$, $\{3,8\}$, $\{4,7\}$ and $\{5,6\}$. Now use pigeon hole principle.
• It's a little tricky in that we are asked to show two pairs have sum $11$, and the pigeonhole principle gives one pair by immediate application (even if as few as six integers are selected). The difference between selecting seven and six integers from $1$ to $10$ is that we get (at least) two pairs with sum $11$ in the former case and as few as one pair in the latter case. – hardmath Dec 1 '13 at 16:58