2014 AMC 10B Problems/Problem 14

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 $a\ge1$ and $a+b+c\le7$ . 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

Let $h\in\mathbb{N}$ be the number of hours Danica drove. Note that $abc$ can be expressed as $100\cdot a+10\cdot b+c$ . From the given information, we have $100a+10b+c+55h=100c+10b+a$ . This can be simplified into $99a+55h=99c$ by subtraction, which can further be simplified into $9a+5h=9c$ by dividing both sides by $11$ . Thus we must have $h\equiv0\pmod9$ . However, if $h\ge 16$ , then $\text{min}\{c\}\ge\frac{9+5(16)}{9}\ge9$ , which is impossible since $c$ must be a digit. The only value of $h$ that is divisible by $9$ and less than or equal to $15$ is $h=9$ .

From this information, $9a+5(9)=9c\Rightarrow a+5=c$ . Combining this with the inequalities $a+b+c\le7$ and $a\ge1$ , we have $a+b+a+5\le7\Rightarrow 2a+b\le2$ , which implies $1\le a\le1$ , so $a=1$ , $b=0$ , and $c=6$ . Thus $a^2+b^2+c^2=1+0+36=\boxed{37 \textbf {(D) }}

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2014 AMC 10B Problems/Problem 14

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