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

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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>.
 
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 ==
 
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{{AMC10 box|year=2007|ab=B|num-b=7|num-a=9}}
 
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{{MAA Notice}}
 

Revision as of 13:06, 5 November 2023

CHIKEN NUGGIEs

Solution 2

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 $8+6+4+2=\boxed{\textbf{(D)}\ 20}$.

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