2014 AMC 10B Problem 14

1. How far did she drive? The odometer went from $abc$ to $cba$. Write each as a number: $abc = 100a + 10b + c$ and $cba = 100c + 10b + a$.
2. Subtract. The middle digit cancels:
$$(100c + 10b + a) - (100a + 10b + c) = 99c - 99a = 99(c - a).$$
(Check with an example: $601 - 106 = 495 = 99 \times 5$.)
3. Use the speed. She drove $h$ whole hours at $55$ miles per hour, so $99(c - a) = 55h$.
4. Both $99$ and $55$ are multiples of $11$. Divide by $11$: $9(c - a) = 5h$.
5. The left side is a multiple of $9$ and the right side is a multiple of $5$. Since $9$ has no factor $5$, the number $c - a$ must be a multiple of $5$.
6. She actually drove, so $c - a$ is positive. It is a difference of two digits, so it is at most $9$. The only positive multiple of $5$ up to $9$ is $5$. So $c = a + 5$ (and $h = 9$ hours).
7. Use the digit limit. $a + b + c \le 7$ becomes $a + b + (a + 5) \le 7$, so $2a + b \le 2$. Since $a \ge 1$, the only choice is $a = 1$, $b = 0$. Then $c = 6$.
8. Check: $601 - 106 = 495 = 55 \times 9$. Then $a^2 + b^2 + c^2 = 1 + 0 + 36 = 37$.

The answer is $\boxed{\textbf{(D)}\ 37}$.

Reversing the digits of a 3-digit number always changes it by a multiple of $99$, because the middle digit stays put. Matching that with a multiple of $55$ leaves only one possible gap between the outer digits. A strange-looking condition like $a + b + c \le 7$ is usually there to pick out exactly one answer at the end.

If the trip lasted $h$ hours, the distance is $55h$, so
$$\overline{cba} - \overline{abc} = (100c + 10b + a) - (100a + 10b + c) = 99(c - a) = 55h.$$

Divide by $11$: $9(c - a) = 5h$. Since $\gcd(9, 5) = 1$, the number $5$ must divide $c - a$. The trip was real ($h \ge 1$), so $c - a > 0$; and as a difference of digits $c - a \le 9$, so $c - a = 5$ (and $h = 9$).

Now use $a \ge 1$ and $a + b + c \le 7$. With $c = a + 5$, the sum is $2a + b + 5 \le 7$, so $2a + b \le 2$. Since $a \ge 1$, this forces $a = 1$ and $b = 0$, hence $c = 6$.

Check: $601 - 106 = 495 = 55 \cdot 9$. Then $a^2 + b^2 + c^2 = 1 + 0 + 36 = 37$.

The answer is $\boxed{\textbf{(D)}\ 37}$.