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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>.
  
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<math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}}\ 37\qquad\textbf{(E)}\ 41 </math>
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<math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41 </math>
  
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==Solution==
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==Solution 1==
  
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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
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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 equal or differ by <math>5</math>. We can quickly conclude that the former is impossible; otherwise, she would have travelled 0 miles. Therefore, <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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==Solution 2==
 +
Let the number of hours Danica drove be <math>k</math>. Then we know that <math>100a + 10b + c + 55k</math> = <math>100c + 10b + a</math>. Simplifying, we have <math>99c - 99a = 55k</math>, or <math>9c - 9a = 5k</math>. Thus, k is divisible by <math>9</math>. Because <math>55 * 18 = 990</math>, <math>k</math> must be <math>9</math>, and therefore <math>c - a = 5</math>. Because <math>a + b + c \leq{7}</math> and <math>a \geq{1}</math>, <math>a = 1</math>, <math>c = 6</math> and <math>b = 0</math>, and our answer is <math>a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37</math>, or <math>\boxed{D}</math>.
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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}}

Latest revision as of 04:00, 2 November 2024

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 equal or differ by $5$. We can quickly conclude that the former is impossible; otherwise, she would have travelled 0 miles. Therefore, $a$ 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 $\boxed{D}$.


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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