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Difference between revisions of "2007 AMC 10B Problems/Problem 8"

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==Problem 8==
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== Problem ==
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On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form <math>bbcac,</math> where <math>0 \le a < b < c \le 9,</math> and <math>b</math> was the average of <math>a</math> and <math>c.</math> How many different five-digit numbers satisfy all these properties?
 
On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form <math>bbcac,</math> where <math>0 \le a < b < c \le 9,</math> and <math>b</math> was the average of <math>a</math> and <math>c.</math> How many different five-digit numbers satisfy all these properties?
  
 
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25</math>
 
<math>\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25</math>
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==Solution==
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Case <math>1</math>: The numbers are separated by <math>1</math>.
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We this case with <math>a=0, b=1,</math> and <math>c=2</math>. Following this logic, the last set we can get is <math>a=7, b=8,</math> and <math>c=9</math>. We have <math>8</math> sets of numbers in this case.
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Case <math>2</math>: The numbers are separated by <math>2</math>.
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This case starts with <math>a=0, b=2,</math> and <math>c=2</math>. It ends with <math>a=5, b=7,</math> and <math>c=9</math>. There are <math>6</math> sets of numbers in this case.
  
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==Solution==
 
  
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For the average, <math>b,</math> to be an integer, <math>a</math> and <math>c</math> have to either both be odd or both be even. There are <math>\frac{5 \times
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Case <math>3</math>: The numbers start with <math>a=0, b=3,</math> and <math>c=6</math>. It ends with <math>a=3, b=6,</math> and <math>c=9</math>. This case has <math>4</math> sets of numbers.
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4}{2 \times 1} = 10</math> ways to choose a set of two even numbers and <math>\frac{5 \times 4}{2 \times 1} = 10</math> ways to choose a set
 
  
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of two odd numbers. Therefore, the number of five-digit numbers that satisfy these properties is <math>10+10=\boxed{\textbf{(D) }20}</math>
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It's pretty clear that there's a pattern: <math>8</math> sets, <math>6</math> sets, <math>4</math> sets. The amount of sets per case decreases by <math>2</math>, so it's obvious Case <math>4</math> has <math>2</math> sets. The total amount of possible five-digit numbers is <math>8+6+4+2=\boxed{\textbf{(D)}\ 20}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2007|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2007|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:00, 5 July 2024

Problem

On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $bbcac,$ where $0 \le a < b < c \le 9,$ and $b$ was the average of $a$ and $c.$ How many different five-digit numbers satisfy all these properties?

$\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25$

Solution

Case $1$: The numbers are separated by $1$.

We this case with $a=0, b=1,$ and $c=2$. Following this logic, the last set we can get is $a=7, b=8,$ and $c=9$. We have $8$ sets of numbers in this case.


Case $2$: The numbers are separated by $2$.

This case starts with $a=0, b=2,$ and $c=2$. It ends with $a=5, b=7,$ and $c=9$. There are $6$ sets of numbers in this case.


Case $3$: The numbers start with $a=0, b=3,$ and $c=6$. It ends with $a=3, b=6,$ and $c=9$. This case has $4$ sets of numbers.

It's pretty clear that there's a pattern: $8$ sets, $6$ sets, $4$ sets. The amount of sets per case decreases by $2$, so it's obvious Case $4$ has $2$ sets. The total amount of possible five-digit numbers is $8+6+4+2=\boxed{\textbf{(D)}\ 20}$.

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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