Question 1127420
{{{ highlight( 18 ) }}}

—

Clearly {{{ N >= 14 }}} <br>

Say the colors are A,B,C,D, and E.<br>

When Peter first starts picking socks from the drawer, the greatest number of socks he can pick without getting a single pair is 5 (A,B,C,D,E).  Upon drawing the 6th sock (A) he is guaranteed to have a pair.   At this point, he either draws an A which he holds unmatched OR he draws another color that pairs up (and hence the pair count goes up by 1).   

<pre>
Worst case pairing table:
Draw:       1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
# pairs:    0, 0, 0, 0, 0, 1, 1, 2, 2,  3,  3,  4,  4,  5,  5,  6,  6,  7

</pre>     

Starting with draw #6, if we assume any color on any draw, then the drawn color either replaces a previously unmatched color or it forms a pair, there is no other option.  

—————

Alternative solution:
What is the least number of pairs he can have with 14 socks drawn?  That would be 5 pairs.   
A,B,C,D,E,  A,B,C,D,E,  A,B,C,D

He would also hold 4 of the 5 colors, unmatched.  Now the next draw (#15) would have to be the one color he is not holding (E) in order to avoid making another pair.  At this point though, the next draw (#16) definitely pairs up, bringing the pair count to 6.  He'd have to then draw the color just paired on #17 in order to avoid making a pair.  Finally,  draw #18 is another must-match situation, bringing the pair count to 7.