Difference between revisions of "2014 AMC 12B Problems/Problem 10"

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<cmath>a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}</cmath>
 
<cmath>a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}</cmath>
  
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== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:23, 22 February 2014

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 $a^2+b^2+c^2?$.

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}}\ 37\qquad\textbf{(E)}\ 41$ (Error compiling LaTeX. Unknown error_msg)

Solution

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 \[a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}\]

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
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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