2014 AMC 12B Problems/Problem 10: Difference between revisions

Revision as of 12:23, 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 $a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}$

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

@@ Line 10: / Line 10: @@
 <cmath>a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{\textbf{(D)}\ 37}</cmath>
+== See also ==
 {{AMC12 box|year=2014|ab=B|num-b=9|num-a=11}}
 {{MAA Notice}}

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2014 AMC 12B Problems/Problem 10: Difference between revisions

Revision as of 12:23, 22 February 2014

Problem

Solution

See also