Difference between revisions of "2014 AMC 12B Problems/Problem 10"
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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, <math>abc</math> miles was displayed on the odometer, where <math>abc</math> is a 3-digit number with <math>a \geq{1}</math> and <math>a+b+c \leq{7}</math>. At the end of the trip, the odometer showed <math>cba</math> miles. What is <math>a^2+b^2+c^2?</math>. | 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, <math>abc</math> miles was displayed on the odometer, where <math>abc</math> is a 3-digit number with <math>a \geq{1}</math> and <math>a+b+c \leq{7}</math>. At the end of the trip, the odometer showed <math>cba</math> miles. What is <math>a^2+b^2+c^2?</math>. | ||
− | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 </math> |
==Solution== | ==Solution== |
Revision as of 09:12, 3 March 2015
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 .
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
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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