2011 AMC 10A Problems/Problem 13

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 1

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.

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2011 AMC 10A Problems/Problem 13

Contents

Problem 13

Solution 1

Solution 2

See Also