Difference between revisions of "2011 AMC 10A Problems/Problem 13"
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==Problem 13== | ==Problem 13== | ||
| − | How many even integers are there between <math>200</math> and <math>700</math> whose digits are all different and come from the set <math>\left\{1,2,5,7,8,9 | + | How many even integers are there between <math>200</math> and <math>700</math> whose digits are all different and come from the set <math>\left\{1,2,5,7,8,9\right\}</math>? |
| − | < | + | <math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math> |
== 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 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. | 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. | ||
| + | |||
| + | == Solution 2== | ||
| + | |||
| + | We see that the last digit of the <math>3</math>-digit number must be even to have an even number. Therefore, the last digit must either be <math>2</math> or <math>8</math>. | ||
| + | |||
| + | 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=8</math> numbers. | ||
| + | |||
| + | Adding up our cases, we have <math>4+8=\boxed{\text{(A)}12}</math> numbers. | ||
== See Also == | == See Also == | ||
Revision as of 18:15, 5 January 2020
Contents
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.
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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