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Difference between revisions of "2014 AMC 10B Problems/Problem 14"
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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>
 
<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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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 15</math>, then <math>\text{min}\{c\}\ge\frac{9+5(15)}{9}>9</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>14</math> is <math>h=9</math>.  
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Let <math>h</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 15</math>, then <math>\text{min}\{c\}\ge\frac{9+5(15)}{9}>9</math>, which is impossible since <math>c</math> must be a digit. The only value of <math>h</math> divisible by <math>9</math> and less than or equal to <math>14</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 <math>a^2+b^2+c^2=1+0+36=\fbox{37 \textbf{(D)}}</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>a=1</math>, so <math>b=0</math>, and <math>c=6</math>. Thus <math>a^2+b^2+c^2=1+0+36=\fbox{37 \textbf{(D)}}</math>
  
 
==Solution 2==
 
==Solution 2==
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This simplifies to be <math>0=55x+99a-99c\implies -55x=99a-99c</math>
 
This simplifies to be <math>0=55x+99a-99c\implies -55x=99a-99c</math>
  
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Factoring, we get that <math>-55x=99(a-c)\implies </math>x=-\frac{9(a-c)}{5}<math>
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Factoring, we get that <math>-55x=99(a-c)\implies x=-\frac{9(a-c)}{5}</math>
  
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Hence, notice that we want </math>a-c=-5<math>
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Hence, notice that we want <math>a-c=-5</math> so that <math>x=9</math>
  
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The only pair that works for this problem that satisfies the original requirements is </math>(1,6)<math>
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The only pair that works for this problem that satisfies the original requirements is <math>(1,6)</math>
  
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Hence, </math>a=1, b=0, c=6<math>
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Hence, <math>a=1, b=0, c=6</math>
  
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Checking, we have that </math>106+495=601<math>
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Checking, we have that <math>106+55(9)=601\implies 601=601</math>
  
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Hence, the answer is </math>1^2+0^2+6^2=37\implies\boxed{D}$
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Hence, the answer is <math>1^2+0^2+6^2=37\implies\boxed{D}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/C0erYBsw5KI
 +
 
 +
~savannahsolver
  
 
==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}}

Latest revision as of 23:13, 17 September 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 $a\ge1$ 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 1

Let $h$ 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 $h\ge 15$, then $\text{min}\{c\}\ge\frac{9+5(15)}{9}>9$, which is impossible since $c$ must be a digit. The only value of $h$ divisible by $9$ and less than or equal to $14$ 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 $a=1$, so $b=0$, and $c=6$. Thus $a^2+b^2+c^2=1+0+36=\fbox{37 \textbf{(D)}}$

Solution 2

Danica drives $m$ miles, such that $m>0$ and $m$ is a multiple of 55. Therefore, $m$ must have an units digit of either $0$ or $5.$ If the units digit of $m$ is $0,$ then $a=c$ which would imply that Danica did not drive at all. Thus, $c>a.$ Therefore, $|a-c|=5,$ and because $a+b+c\leq7, c>a,$ we have $(a,c)=(1,6).$ Finally, $b$ then must be $0$ due to $a+b+c\leq 7,$ and $a^2+b^2+c^2=1^2+0^2+6^2=\fbox{\textbf{(D) }37}$

Solution 3

We can set up an algebraic equation for this problem.

From what's given, we have that $100c+10b+a=55x+100a+10b+c$

This simplifies to be $0=55x+99a-99c\implies -55x=99a-99c$

Factoring, we get that $-55x=99(a-c)\implies x=-\frac{9(a-c)}{5}$

Hence, notice that we want $a-c=-5$ so that $x=9$

The only pair that works for this problem that satisfies the original requirements is $(1,6)$

Hence, $a=1, b=0, c=6$

Checking, we have that $106+55(9)=601\implies 601=601$

Hence, the answer is $1^2+0^2+6^2=37\implies\boxed{D}$

Video Solution

https://youtu.be/C0erYBsw5KI

~savannahsolver

See Also

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

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