# Difference between revisions of "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

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 $1 \times 4 \times 2=8$ 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.

## See Also

 2011 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 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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