The solution to A2 of Putnams 2006 claims that for any integer $\notin B$ it must be of the form $b+p-1$ for some $b\in B$ and prime $p$ which is clear so far.
I'm confused about the existence of the integer $x$ according to the chinese remainder theorem. Why is any integer of the form $b+p-1$ a solution of the given simultaneous congruence?
Anyone mind elaborating?
Thanks
Here is a link to the 2006 Putnam: http://math.hawaii.edu/home/pdf/putnam/2006.pdf
An alternative proof: Suppose set $S$ of second-player-wins positions is finite, hence bounded by some $m$. Then for every $n>m$, there must exist a move taking $n$ to $S$, i.e., a prime $p$ with $0\le n-(p-1)\le m$. In other words, there must be a prime $p\in\{n-m+1,\ldots, n+1\}$, a set of $m+1$ consecutive integers. However, the existence of arbitraryly large prime gaps is clear, e.g., each of the $N-1$ consecutive integers $N!+2,\ldots,N! +N$ is composite (for $N\ge 2$). So concretely, $n=(m+2)!+m+1$ must be $\in S$ while being $>m$.