2014 AMC 12B Problems/Problem 10

Problem

Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$ .

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41$

Solution 1

We know that the number of miles she drove is divisible by $5$ , so $a$ and $c$ must either be equal or differ by $5$ . We can quickly conclude that the former is impossible; otherwise, she would have travelled 0 miles. Therefore, $a$ and $c$ must be $5$ apart. Because we know that $c > a$ and $a + c \le 7$ and $a \ge 1$ , we find that the only possible values for $a$ and $c$ are $1$ and $6$ , respectively. Because $a + b + c \le 7$ , $b = 0$ . Therefore, we have $a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}$

Solution 2

Let the number of hours Danica drove be $k$ . Then we know that $100a + 10b + c + 55k$ = $100c + 10b + a$ . Simplifying, we have $99c - 99a = 55k$ , or $9c - 9a = 5k$ . Thus, k is divisible by $9$ . Because $55 * 18 = 990$ , $k$ must be $9$ , and therefore $c - a = 5$ . Because $a + b + c \leq{7}$ and $a \geq{1}$ , $a = 1$ , $c = 6$ and $b = 0$ , and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$ , or $\boxed{D}$ .

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

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2014 AMC 12B Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See also