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2014 AMC 12B Problems/Problem 10

Revision as of 02:04, 16 January 2021 by Joeya (talk | contribs)

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 the equal or differ by $5$. We can quickly conclude that the former is impossible, so $a$ 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 $\boxed{D}$.

Solution 3 (Testing Values)

We know that since adding a multiple of $55$ will increase the number, we know that $a \leq{c}$. After testing values for $a$, $b$, and $c$, adding their squares, we see that using non-zero integers as the digits can yield a maximum value of $a^2 + b^2 + c^2$ as 27, specifically the combination $(a, b, c)$ of $(1, 1, 5)$. However, we see that no multiple of $55$ can be added to $115$ to achieve $511$. No combination of non-zero integers can create a higher answer, so we must try testing with a zero. Realizing that only $b$ can be zero, we first try to see if $(1, 0, 5)$ (which creates 26 from the expression $1^2 + 0^2 + 5^2$, answer choice A) can be added to a multiple of $55$ to create $501$, which we find it cannot. However, we quickly realize that the next consecutive pair $(1, 0, 6)$ does satisfy this condition, and $1^2 + 0^2 + 6^2 = \boxed{\textbf{(D)}\ 37}$. We come to this quickly after realizing that if the multiple of $55$ was an even multiple, it would have to end in $0$ and thus the $c$ digit would remain unchanged, so it must be an odd multiple, which will carry over the $c$ digit, so $c \geq{6}$.

-Solution by Joeya

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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