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Difference between revisions of "2014 AMC 10B Problems/Problem 14"

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==Solution==
 
==Solution==
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Let <math>h\in\mathbb{N}</math> be the number of hours Danica drove. Note that <math>abc</math> can be expressed as <math>100\cdot a+10\cdot b+c</math>. From the given information, we have <math>100a+10b+c+55h=100c+10b+a</math>. This can be simplified into <math>99a+55h=99c</math> by subtraction, which can further be simplified into <math>9a+5h=9c</math> by dividing both sides by <math>11</math>. Thus we must have <math>h\equiv0\pmod9</math>. However, if <math>h\ge 16</math>, then <math>\text{min}\{c\}\ge\frac{9+5(16)}{9}\ge9</math>, which is impossible since <math>c</math> must be a digit. The only value of <math>h</math> that is divisible by <math>9</math> and less than or equal to <math>15</math> is <math>h=9</math>.
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From this information, <math>9a+5(9)=9c\Rightarrow a+5=c</math>. Combining this with the inequalities <math>a+b+c\le7</math> and <math>a\ge1</math>, we have <math>a+b+a+5\le7\Rightarrow 2a+b\le2</math>, which implies <math>1\le a\le1</math>, so <math>a=1</math>, <math>b=0</math>, and <math>c=6</math>. Thus $a^2+b^2+c^2=1+0+36=\boxed{37 \textbf {(D) }}
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2014|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:44, 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, $abc$ miles was displayed on the odometer, where $abc$ is a $3$-digit number with $a\ge1$ and $a+b+c\le7$. 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

Let $h\in\mathbb{N}$ be the number of hours Danica drove. Note that $abc$ can be expressed as $100\cdot a+10\cdot b+c$. From the given information, we have $100a+10b+c+55h=100c+10b+a$. This can be simplified into $99a+55h=99c$ by subtraction, which can further be simplified into $9a+5h=9c$ by dividing both sides by $11$. Thus we must have $h\equiv0\pmod9$. However, if $h\ge 16$, then $\text{min}\{c\}\ge\frac{9+5(16)}{9}\ge9$, which is impossible since $c$ must be a digit. The only value of $h$ that is divisible by $9$ and less than or equal to $15$ is $h=9$.

From this information, $9a+5(9)=9c\Rightarrow a+5=c$. Combining this with the inequalities $a+b+c\le7$ and $a\ge1$, we have $a+b+a+5\le7\Rightarrow 2a+b\le2$, which implies $1\le a\le1$, so $a=1$, $b=0$, and $c=6$. Thus $a^2+b^2+c^2=1+0+36=\boxed{37 \textbf {(D) }}

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 10 Problems and Solutions

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