Difference between revisions of "2006 AMC 12B Problems/Problem 9"
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== Solution == | == 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 <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?
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. Unknown error_msg).
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
