Difference between revisions of "2011 AMC 10A Problems/Problem 13"

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Problem 13

How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$ ?

$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$

Solution

We split up into cases of the hundreds digits being $2$ or $5$ . If the hundred digits is $2$ , then the units digits must be $8$ in order for the number to be even and then there are $4$ remaining choices ( $1,5,7,9$ ) for the tens digit, giving $1 \times 1 \times 4=4$ possibilities. Similarly, there are $1 \times 2 \times 4=8$ possibilities for the $5$ case, giving a total of $\boxed{4+8=12 \ \mathbf{(A)}}$ possibilities.

Solution 2

We see that the last digit of the $3$ -digit number must be even to have an even number. Therefore, the last digit must either be $2$ or $8$ .

Case $1$ -the last digit is $2$ . We must have the hundreds digit to be $5$ and the tens digit to be any $1$ of ${1,7,8,9}$ , thus obtaining $4$ numbers total.

Case $2$ -the last digit is $8$ . We now can have $2$ or $5$ to be the hundreds digit, and any choice still gives us $4$ choices for the tens digit. Therefore, we have $2 \cdot 4=8$ numbers.

Adding up our cases, we have $4+8=\boxed{\text{(A)}12}$ numbers.

Solution 3 (elimination)

We see that there are $\dbinom63=20$ total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set ${1,2,5,7,8,9}.$ Obviously, some of these (such as $987,$ for example) will not work, and thus the answer will be less than $20.$ The only possible option is $\boxed{\text{(A)} 12}.$ ~ Technodoggo

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 == 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 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.
+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.
 == Solution 2==

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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