Math Parlor Trick 
A magician asks a person in the audience to think of a number $\overline {abc}$. He then asks them to sum up $\overline{acb}, \overline{bac}, \overline{bca}, \overline{cab}, \overline{cba}$ and reveal the result. Suppose it is $3194$. What was the original number?

The obvious approach was modular arithmetic. 
$(100a + 10c + b) + (100b + 10a + c) + (100b + 10c + a) + (100c + 10a + b) + (100c + 10b + a) = 3194$
$122a + 212b + 221c = 3194$
Since $122, 212, 221 \equiv 5 (mod\space9)$ and $3194 \equiv 8 (mod\space9)$
$5(a + b + c) \equiv 8 (mod\space9)$
So, $a + b + c = 7$ or $16$ or $26$
Hit and trial produces the result $358$. Any other, more elegant method?

Let $S$ be the sum,
$$S \text{ mod} 10=A$$
$$S \text{ mod} 100=B$$
$$A=2b+2a+c$$
$$\frac{B-A}{10}=(2c+2a+b)$$
$$\frac{S-B}{100}=(a+2b+2c)$$
$$\text{Now just solve the system of equations for $a$ $b$ and $c$}$$
$$\text{ The original number will be a+10b+100c}$$
$$\text{ Now memorize this formula and do the addition in your head}$$
