Difference between revisions of "2011 AMC 10A Problems/Problem 13"
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| − | We split up into cases of the hundreds digits being <math>2</math> or <math>5</math>. If the hundred digits is <math>2</math>, then the units digits must be <math>8</math> in order for the number to be even and then there are <math>4</math> remaining choices (<math>1,5,7,9</math>) for the tens digit, giving <math>1 \times 1 \times 4=4</math> possibilities. Similarly, there are <math>1 \times 2 \times 4=8</math> possibilities for the <math>5</math> case, giving a total of \boxed{12 \ \mathbf{(A)}} possibilities. | + | We split up into cases of the hundreds digits being <math>2</math> or <math>5</math>. If the hundred digits is <math>2</math>, then the units digits must be <math>8</math> in order for the number to be even and then there are <math>4</math> remaining choices (<math>1,5,7,9</math>) for the tens digit, giving <math>1 \times 1 \times 4=4</math> possibilities. Similarly, there are <math>1 \times 2 \times 4=8</math> possibilities for the <math>5</math> case, giving a total of <math>\boxed{4+8=12 \ \mathbf{(A)}}</math> possibilities. |
Revision as of 14:10, 14 February 2011
Problem 13
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?
Solution
We split up into cases of the hundreds digits being 
or 
. If the hundred digits is , then the units digits must be 
in order for the number to be even and then there are 
remaining choices (
) for the tens digit, giving 
possibilities. Similarly, there are 
possibilities for the case, giving a total of 
possibilities.
