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

(Problem 13)
m
 
(3 intermediate revisions by one other user not shown)
Line 7: Line 7:
  
 
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 4 \times 1=4</math> possibilities. Similarly, there are <math>1 \times 4 \times 2=8</math> possibilities for the <math>5</math> case, giving a total of <math>\boxed{4+8=12 \ \mathbf{(A)}}</math> possibilities.
 
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 4 \times 1=4</math> possibilities. Similarly, there are <math>1 \times 4 \times 2=8</math> possibilities for the <math>5</math> case, giving a total of <math>\boxed{4+8=12 \ \mathbf{(A)}}</math> possibilities.
 +
 +
== Solution 2==
 +
 +
We see that the last digit of the <math>3</math>-digit number must be even to have an even number. Therefore, the last digit must either be <math>2</math> or <math>8</math>.
 +
 +
Case <math>1</math>-the last digit is <math>2</math>. We must have the hundreds digit to be <math>5</math> and the tens digit to be any <math>1</math> of <math>{1,7,8,9}</math>, thus obtaining <math>4</math> numbers total.
 +
 +
Case <math>2</math>-the last digit is <math>8</math>. We now can have <math>2</math> or <math>5</math> to be the hundreds digit, and any choice still gives us <math>4</math> choices for the tens digit. Therefore, we have <math>2 \cdot 4=8</math> numbers.
 +
 +
Adding up our cases, we have <math>4+8=\boxed{\text{(A)}12}</math> numbers.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:15, 5 January 2020

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 (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

Invalid username
Login to AoPS