Difference between revisions of "2011 AMC 10A Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | 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 | + | 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 4 \times 1=4</math> possibilities. Similarly, there are <math>1 \times 4 \times 2=8</math> possibilities for the <math>5</math> case, giving a total of <math>\boxed{4+8=12 \ \mathbf{(A)}}</math> possibilities. |
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== See Also == | == See Also == | ||
Revision as of 15:46, 18 August 2018
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.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 10 Problems and Solutions | ||
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