for part (a) abby has the winning strategy, he must always write an odd number on the blackboard.

To rephrase: a) First, note that Abby writes an odd number, so that Brian must write an even number. After an even number, Abby can always write an odd number (n+1), which forces Brian to write an even number. Since 2011 is odd, and since Brian can be forced not to write an odd number, then Abby wins. For b, my first step is to use the above argument to prove that N must be even, N=2 works, and N=4, N=6 fails. Then...well, I'm not able to find the next step as of now.

b)There are ${2^5} = 32$ positive integers $N<2012$ for which Brian has a winning strategy.

yunxiu wrote: b)There are ${2^5} = 32$ positive integers $N<2012$ for which Brian has a winning strategy. That is not correct, the answer has to be $31$ so without explaining you are completely wrong. I think the solution is like this; It is easy that if $B$ can win for $n$ he can also win for $4n$ and $4n+2$ $A\to n+1$ gives $B\to 2n+2$ $A\to 2n,B\to 4n$ and only adding $1$ per person, so $B$ holds all even numbers and will win. It was trivially $A$ wins the odd numbers by $(a)$ [ but you had to do it by (b) or you got only 1/7] and so $A$ wins $4n+1$ and $4n+3$ $A$ can win all other games, because he has the $2$ odd numbers and for other $n$ he wins $4n$ until $4n+3.$ So starting with $n=2$ which is first number that $B$ can win, and easy $A$ wins $3,4,5,6,7$ We see $B$ wins $8,10$ A: $9,11,12,13,\cdots,31$ B: $32,34,40,42$ and $A$ other ones until $127$ and so on and you get $31.$ From each $n$ you go to $4n,4n+2$ and so $2$ new ones. Because $4^5n>2011$ because $n=2$ to start, we can't do more than $2^4$ at last and $1+2+\cdots+2^4=31.$

SCP wrote: yunxiu wrote: b)There are ${2^5} = 32$ positive integers $N<2012$ for which Brian has a winning strategy. That is not correct, the answer has to be $31$ I am sorry, you are right. $2011={\left( {\overline {11111011011} } \right)_2}$ , $N\leqslant 2011$ for which Brian has a winning strategy iff $N={\left( {\overline {0{a_9}0{a_7}0{a_5}0{a_3}0{a_1}0} } \right)_2}$， where ${a_1},{a_3},{a_5},{a_7},{a_9}=0 or 1$, but $N \ne 0$. So there are ${2^5} - 1 = 31$ positive integers for which Brian has a winning strategy.