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

m (Solution 1)
(Solution 3 (elimination))
Line 21: Line 21:
 
We see that there are <math>\dbinom63=20</math> total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set <math>{1,2,5,7,8,9}.</math> Obviously, some of these (such as <math>987,</math> for example) will not work, and thus the answer will be less than <math>20.</math> The only possible option is <math>\boxed{\text{(A)} 12}.</math>
 
We see that there are <math>\dbinom63=20</math> total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set <math>{1,2,5,7,8,9}.</math> Obviously, some of these (such as <math>987,</math> for example) will not work, and thus the answer will be less than <math>20.</math> The only possible option is <math>\boxed{\text{(A)} 12}.</math>
 
~ Technodoggo
 
~ Technodoggo
 +
 +
*This doesn't make sense because there are <math>6 * 5 * 4</math> to make a 3-digit number from the given set of 6 numbers.
  
 
== See Also ==
 
== See Also ==

Revision as of 13:44, 3 November 2024

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.

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

  • This doesn't make sense because there are $6 * 5 * 4$ to make a 3-digit number from the given set of 6 numbers.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png