Mathematical Puzzles: A & B erase lines from an array of N lines, taking turns. If A begins erasing lines, what strategy must he/she adopt in order to ensure that he/she does not have to erase the last set of lines?

The proof follows quite easily from the observation that whenever A plays, A should strive to leave [math]b_0[/math]mod[math](b_1 + a_0)[/math] lines in the array. Let's call this the b1a0-post-condition. This ensures that there will come a time when only [math]b_0[/math] lines will be left and B will be forced to erase them all. Lemmas to be proved, left as an exercise for the reader: If on the first step, A erases [math](N - b_0)[/math]mod[math](b_1 + a_0)[/math] then the A-post-condition is satisfied If the b1a0-post-condition is satisfied, and B erases [math]x[/math], then for any [math]x[/math] the b1a0-post-condition is no longer satisfied. At this point when A erases [math]b_1 + a_0 - x[/math] the b1a0-post-condition is satisfied. At any point, if b1a0-post-condition is satisfied, B can never erase enough lines so that less than or equal to [math]a_0[/math] lines are left and A is forced to pick them all up. For extra credit: Figure out what happens if [math]N = b_0 mod (b_1 + a_0)[/math]. What are the conditions under which A can still guarantee a win? Also, please, just reframe this question in terms of [math]N[/math] coins on a table, and A and B taking turns to pick up some number of coins. It's much easier to explain and reason about. Don't get unnecessarily nerdy.