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

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<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math>
 
<math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200</math>
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== Solution ==
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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 \boxed{12 \ \mathbf{(A)}} possibilities.

Revision as of 14:08, 14 February 2011

Problem 13

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

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

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{12 \ \mathbf{(A)}} possibilities.