Difference between revisions of "2014 AMC 12B Problems/Problem 10"
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We know that the number of miles she drove is divisible by <math>5</math>, so <math>a</math> and <math>c</math> must either be the equal or differ by <math>5</math>. We can quickly conclude that the former is impossible, so <math>a</math> and <math>c</math> must be <math>5</math> apart. Because we know that <math>c > a</math> and <math>a + c \le 7</math> and <math>a \ge 1</math>, we find that the only possible values for <math>a</math> and <math>c</math> are <math>1</math> and <math>6</math>, respectively. Because <math>a + b + c \le 7</math>, <math>b = 0</math>. Therefore, we have | We know that the number of miles she drove is divisible by <math>5</math>, so <math>a</math> and <math>c</math> must either be the equal or differ by <math>5</math>. We can quickly conclude that the former is impossible, so <math>a</math> and <math>c</math> must be <math>5</math> apart. Because we know that <math>c > a</math> and <math>a + c \le 7</math> and <math>a \ge 1</math>, we find that the only possible values for <math>a</math> and <math>c</math> are <math>1</math> and <math>6</math>, respectively. Because <math>a + b + c \le 7</math>, <math>b = 0</math>. Therefore, we have | ||
<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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Revision as of 21:36, 20 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, miles was displayed on the odometer, where is a 3-digit number with and . At the end of the trip, the odometer showed miles. What is .
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}}\ 37\qquad\textbf{(E)}\ 41$ (Error compiling LaTeX. Unknown error_msg)
Solution
We know that the number of miles she drove is divisible by , so and must either be the equal or differ by . We can quickly conclude that the former is impossible, so and must be apart. Because we know that and and , we find that the only possible values for and are and , respectively. Because , . Therefore, we have