Sum of the jumbled digits of $abc_{10}$ is $3194$ In the book that I am reading, the author denotes $abc_{10}$ as $100a+10b+c$ where $a, b, c \lt 10$. So if $a = 3$, $b=2$ and $c=8$ then $abc_{10} = 328$.
The author asks the following problem: 
In a parlor game, the 'magician' asks one of the participants to think of a three-digit number $abc_{10}$. The the magician asks the participant to add the five numbers $acb_{10}$, $bac_{10}$, $bca_{10}$, $cab_{10}$ and $cba_{10}$. Suppose the sum was $3194$. What was $abc_{10}$ originally?
I changed the five numbers to $100a+10b+c$ form and added the five numbers that way, resulting in the equation $122a+212b+221c=3194$. Or you could add them another way and factor the result to make $100(a+2b+2c)+10(2a+b+2c)+1(2a+b+c)$. 
I am not sure what to do next, but the author hinted at using modular arithmetic to get bounds on the three variables. The only thing I can get with modular arithmetic is the following equation:$$2a+2b+c \equiv 4 \pmod{10} $$ I also found that $c$ is an even number.
Thank you for any help.
Note: If anyone has a better title, please tell me.
 A: If you add all six numbers $abc_{10}+acb_{10}+bac_{10}+bca_{10}+cab_{10}+cba_{10}$, you should obtain the single number
$$
(2a+2b+2c)(2a+2b+2c)(2a+2b+2c)_{10} = 111(2a+2b+2c) = 222(a+b+c)
$$
Therefore,
$$
3194+abc_{10} = 222(a+b+c)
$$
We find that $3194 \bmod 222 = 86$, so $abc_{10}$ must be $222k-86$ for some positive integer $k$.  Trial and error: We try $k = 1$ and find that
$$
3194+136 = 3330 = 222(15)
$$
but $1+3+6 = 10 \not= 15$.  However, $k = 2$ yields
$$
3194+358 = 3552 = 222(16)
$$
and $3+5+8 = 16$, so $abc_{10} = 358$.  By way of confirmation, $385+538+583+835+853 = 3194$.  (I leave the exercise of determining whether there are other solutions to the reader.)
A: Imagine that we accumulate without carries in each value place. Then the digit total of $acb+bac+bca+cab+cba$ would be divisible by $5$. However each time we have a carry operation, that digit value reduces by $9$. 
Now the lower limit on what the pre-carry sum was is $31+9+4 = 44$, because there weren't any "native"  thousands digits going in to the sum. Now we can add the "lost" $9$s back into that to find a multiple of $5$, getting to $80$ (which gives $a+b+c = 16$) and then $125$ (which gives $a+b+c=25$) as our two feasible initial digit sums. 
We also know that the last digit after adding $abc$ will be even, so $c$ is even. With $a+b+c=25$, the target digit would be $0$ (since $2\cdot 25=50$) and $c=6$ but this doesn't work with $a+b+c=25$. So now we know that $a+b+c=16$ and thus $\fbox{c=8}$ to reach $2$ in the units digit of the final sum to match $2\cdot 16=32$.
Now we can subtract out the $32-8=24$ and then divide by $10$ to get $ac+ba+bc+ca+cb = 317$. Since $a+b=8$ and we need $ab\equiv 3 \bmod 16$ with $b$ odd, we can check $\{17, 35, 53, 71\}$ to find that we need $ab_{10}=35$ and thus $\fbox{$abc_{10} = 358$}$
