2014 AMC 10B Problem 14

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 $\boxed{\textbf{(D)}\ 37}$.

Reversing digits is a classic setup: the middle digit cancels and the difference is $99$ times the difference of the outer digits. Combining that with the divisibility from $55h$ leaves a single Diophantine condition, and the digit-sum bound is there precisely to pin down the unique solution. When a problem gives a seemingly arbitrary bound like $a + b + c \le 7$, expect it to be the final filter.

• Writing cba - abc = 99(c - a) = 55h but then picking any digits with a + b + c <= 7, missing that 9(c - a) = 5h forces c - a = 5 and b = 0.
• Forgetting that $b$ must be $0$ (choosing $b = 1$ violates the digit-sum bound) or forgetting $a \ge 1$ and trying $a = 0$.