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 4 \times 1=4</math> possibilities. Similarly, there are <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 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. |
== Solution 2== | == Solution 2== | ||
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Case <math>1</math>-the last digit is <math>2</math>. We must have the hundreds digit to be <math>5</math> and the tens digit to be any <math>1</math> of <math>{1,7,8,9}</math>, thus obtaining <math>4</math> numbers total. | Case <math>1</math>-the last digit is <math>2</math>. We must have the hundreds digit to be <math>5</math> and the tens digit to be any <math>1</math> of <math>{1,7,8,9}</math>, thus obtaining <math>4</math> numbers total. | ||
| − | Case <math>2</math>-the last digit is <math>8</math>. We now can have <math>2</math> or <math>5</math> to be the hundreds digit, and any choice still gives us <math>4</math> choices for the tens digit. Therefore, we have <math>2 \cdot 4</math> numbers. | + | Case <math>2</math>-the last digit is <math>8</math>. We now can have <math>2</math> or <math>5</math> to be the hundreds digit, and any choice still gives us <math>4</math> choices for the tens digit. Therefore, we have <math>2 \cdot 4=8</math> numbers. |
| − | Adding up our cases, we have <math>4+8=\boxed{\text{(A)}12</math> numbers. | + | Adding up our cases, we have <math>4+8=\boxed{\text{(A)}12}</math> numbers. |
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| + | ==Solution 3 (elimination)== | ||
| + | We see that there are <math>\dbinom63=20</math> total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set <math>{1,2,5,7,8,9}.</math> Obviously, some of these (such as <math>987,</math> for example) will not work, and thus the answer will be less than <math>20.</math> The only possible option is <math>\boxed{\text{(A)} 12}.</math> | ||
| + | ~ Technodoggo | ||
== See Also == | == See Also == | ||
Latest revision as of 17:50, 10 August 2023
Problem 13
How many even integers are there between 
and 
whose digits are all different and come from the set 
?
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.
Solution 2
We see that the last digit of the 
-digit number must be even to have an even number. Therefore, the last digit must either be or .
Case 
-the last digit is . We must have the hundreds digit to be and the tens digit to be any of 
, thus obtaining numbers total.
Case -the last digit is . We now can have or to be the hundreds digit, and any choice still gives us choices for the tens digit. Therefore, we have 
numbers.
Adding up our cases, we have 
numbers.
Solution 3 (elimination)
We see that there are 
total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set 
Obviously, some of these (such as 
for example) will not work, and thus the answer will be less than 
The only possible option is 
~ Technodoggo
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
