smc

Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had. After $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents. After $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3b-c$ cents, and $C$ has $4c$ cents. After $C$ gave his money away, $A$ has $4a-4b-4c$ cents, $B$ has $-2a+6b-2c$ cents, and $C$ has $-a-b+7c$ cents. Since all of them have $16$ cents in the end, we can write a system of equations. $$4a-4b-4c=16$$ $$-2a+6b-2c=16$$ $$-a-b+7c=16$$ Note that adding the three equation yields $a+b+c=48$, so $4a+4b+4c=192$. Therefore, $8a=208$, so $a=26$. Solving for $a$ can also be done traditionally. Thus, $A$ started out with $26$ cents, which is answer choice $\boxed{26}$.