Question 1188391
<pre>
Peter's object is to leave Paul with 8 sticks the last time.  Because Paul
can't then pick them all up.  So regardless of how many Paul removes, Peter
will be able to remove the rest, for there will be 7 or less.

So to leave Paul with 8 the last time Peter must have left him with 16
before that. For instance, if Pau1 removes 1, leaving Peter with 15, Peter will
then remove 7 and leave Paul with 8.  Or, say, if Pau1 picks up 7, leaving Peter
with 9, Peter will then pick up 1, leaving Paul with 8. 

So to win, Peter must leave Paul with a multiple of 8 each time.  Peter can
always do that if he starts first.  Beginning with 60 sticks, Peter must leave
Paul with 56, the largest multiple of 8 not exceeding 60.  So Peter must begin
by first removing k=4 sticks, to leave Paul with 56.

[BTW, if Paul starts first, and doesn't know the trick, then if Paul at any time
leaves Peter with anything other than a multiple of 8, Peter can then win
because all he has to do is remove enough to leave Paul with a multiple of 8.]

Edwin</pre>