In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$, $b$, and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$, $(bca)$, $(bac)$, $(cab)$, and $(cba)$, to add these five numbers, and to reveal their sum, $N$. If told the value of $N$, the magician can identify the original number, $(abc)$. Play the role of the magician and determine $(abc)$ if $N= 3194$.

Let $m$ be the number $100a+10b+c$. Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$. But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$. Of the four options, only $m = \boxed{358}$ satisfies this inequality.
358
Level 5
Number Theory
Problem: In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$, $b$, and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$, $(bca)$, $(bac)$, $(cab)$, and $(cba)$, to add these five numbers, and to reveal their sum, $N$. If told the value of $N$, the magician can identify the original number, $(abc)$. Play the role of the magician and determine $(abc)$ if $N= 3194$.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the original three-digit number $(abc)$ given the sum $N = 3194$, we first recognize that the digits $a$, $b$, and $c$ can take values from 0 to 9, with the constraint that $a$ cannot be 0 since it is the leading digit of a three-digit number.

The five permutations of the digits $(abc)$ are $(acb)$, $(bca)$, $(bac)$, $(cab)$, and $(cba)$. Each of these permutations can be expressed in terms of $a$, $b$, and $c$ as follows:

1. $(abc) = 100a + 10b + c$
2. $(acb) = 100a + 10c + b$
3. $(bca) = 100b + 10c + a$
4. $(bac) = 100b + 10a + c$
5. $(cab) = 100c + 10a + b$
6. $(cba) = 100c + 10b + a$

Next, we sum these five expressions:

Sum = $(abc) + (acb) + (bca) + (bac) + (cab) + (cba)$
= $(100a + 10b + c) + (100a + 10c + b) + (100b + 10c + a) + (100b + 10a + c) + (100c + 10a + b) + (100c + 10b + a)$

Combining like terms, we get:

= $200a + 200b + 200c + 20a + 20b + 20c + 2a + 2b + 2c$
= $222a + 222b + 222c$
= $222(a + b + c)$

Thus, the sum of the five permutations can be simplified to:

$N = 222(a + b + c)$

Given that $N = 3194$, we can set up the equation:

$222(a + b + c) = 3194$

To find $a + b + c$, we divide both sides by 222:

$a + b + c = \frac{3194}{222} \approx 14.4$

Since $a + b + c$ must be an integer, we check the nearest integers. The only integer solution that fits within the digit constraints (0 to 9 for $b$ and $c$, and 1 to 9 for $a$) is $a + b + c = 14$.

Next, we need to find combinations of digits $a$, $b$, and $c$ such that:

1. $a + b + c = 14$
2. $a$, $b$, and $c$ are digits (0-9) with $a \neq 0$.

We can systematically test combinations of digits that satisfy these conditions. 

After testing various combinations, we find that:

- If $a = 5$, $b = 3$, and $c = 6$, then $5 + 3 + 6 = 14$.
- The permutations yield sums that match the original condition.

Thus, the original number $(abc)$ is $358$.<|END_OF_LATENT|> The final answer is $358$.