Difference between revisions of "2006 AMC 12B Problems/Problem 9"

(Problem)
(Solution)
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== Solution ==
 
== Solution ==
{{solution}}
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Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 8, the ones digit cannot be even).
 +
 
 +
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.
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If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit.
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If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
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and so on.
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 +
So, the answer is <math>3(1+2)+2(3+4)+1(5+6)=\boxed{34} \RIghtarrow B</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}

Revision as of 14:38, 30 November 2008

Problem

How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?

$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$

Solution

Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 8, the ones digit cannot be even).

If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,

and so on.

So, the answer is $3(1+2)+2(3+4)+1(5+6)=\boxed{34} \RIghtarrow B$ (Error compiling LaTeX. ! Undefined control sequence.).

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 12 Problems and Solutions
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