, 2016

Solution:
Consider $M$ in binary. Assume we start with $M = 0$, then add $a$ to $M$, then add $2b$ to $M$, then add $4c$ to $M$, and so on. After the first addition, the first bit (defined as the rightmost bit) of $M$ is toggled with probability $\frac{1}{2}$. After the second addition, the second bit of $M$ is toggled with probability $\frac{1}{2}$. After the third addition, the third bit is toggled with probability $\frac{1}{2}$, and so on for the remaining three additions. As such, the six bits of $M$ are each toggled with probability $\frac{1}{2}$ — specifically, the $k^{th}$ bit is toggled with probability $\frac{1}{2}$ at the $k^{th}$ addition, and is never toggled afterwards. Therefore, each residue from $0$ to $63$ has probability $\frac{1}{64}$ of occurring, so they are all equally likely. The expected value is then just $\frac{63}{2}$.