Difference between revisions of "2007 AMC 10B Problems/Problem 8"
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| − | ==Problem | + | == 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> | ||
| + | ==Solution== | ||
| − | == | + | 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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| + | 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. | ||
| − | + | 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>. | |
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== 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 
where 
and 
was the average of 
and 
How many different five-digit numbers satisfy all these properties?
Solution
Case 
: The numbers are separated by .
We this case with 
and 
. Following this logic, the last set we can get is 
and 
. We have 
sets of numbers in this case.
Case 
: The numbers are separated by .
This case starts with 
and . It ends with 
and . There are 
sets of numbers in this case.
Case 
: The numbers start with 
and 
. It ends with 
and . This case has 
sets of numbers.
It's pretty clear that there's a pattern: sets, sets, sets. The amount of sets per case decreases by , so it's obvious Case has sets. The total amount of possible five-digit numbers is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
