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

(Problem 13)
(Problem 13)
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==Problem 13==
 
==Problem 13==
How many even integers are there between <math>200</math> and <math>700</math> whose digits are all different and come from the set <math>\left\{1,2,5,7,8,9}\right\$?
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How many even integers are there between <math>200</math> and <math>700</math> whose digits are all different and come from the set <math>\left\{1,2,5,7,8,9\right\}</math>?
  
</math>\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$
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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>
  
 
== Solution ==
 
== Solution ==

Revision as of 22:26, 25 August 2019

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.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 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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