Power-of-two ball game where taking the last loses

An urn starts with $100$ balls. Alice and Bob alternate turns; on each turn a player removes $2^k$ balls for some $k\ge0$ of their choosing ($1,2,4,8,\dots$), provided that many remain. The player forced to take the final ball loses. Alice decides whether she moves first ($p=1$) or second ($p=2$). Suppose Bob removes $32$ balls on his first turn; let $b$ be the largest number Alice can take in reply while still keeping a winning strategy. Compute $100p+b$.

$264$. Powers of two are never multiples of $3$, so a player facing a count $\equiv1\pmod 3$ always hands back a count $\not\equiv1$, and the opponent can restore $\equiv1$ (with $1$ ball left being a loss). Since $100\equiv1$, Alice goes second ($p=2$). After Bob's $32$, $68$ remain and she must leave a count $\equiv1\pmod3$; the largest power of two that does this is $64$, leaving $4$. So $200+64=264$.