Difference between revisions of "2007 AMC 12A Problems/Problem 16"

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== See also ==
 
== See also ==
 +
[[2005 AMC 12A Problems/Problem 11|similar problem]]
 
{{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 08:40, 1 July 2008

Problems

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

$\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256$

Solution

We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.

Common difference Sequences possible Number of sequences
1 $012, \ldots, 789$ 8
2 $024, \ldots, 579$ 6
3 $036, \ldots, 369$ 4
4 $048, \ldots, 159$ 2

This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$.

However, we note that the conditions of the problem require two digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.

See also

similar problem

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions
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