# 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 $1 \times 4 \times 1=4$ 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