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Difference between revisions of "1993 AIME Problems/Problem 1"
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How many even [[integer]]s between 4000 and 7000 have four different digits?  
 
How many even [[integer]]s between 4000 and 7000 have four different digits?  
  
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== Solution ==
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== Solution 1 ==
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The thousands digit is <math>\in \{4,5,6\}</math>. If the thousands digit is even (<math>4,\ 6</math>, 2 possibilities), then there are only <math>\frac{10}{2} - 1 = 4</math> possibilities for the units digit. This leaves <math>8</math> possible digits for the hundreds and <math>7</math> for the tens places, yielding a total of <math>2 \cdot 4 \cdot 8 \cdot 7 = 448</math>.
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The thousands digit is <math>\in \{4,5,6\}</math>.  
  
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If the thousands digit is odd (<math>5</math>, one possibility), then there are <math>5</math> choices for the units digit, with <math>8</math> digits for the hundreds and <math>7</math> for the tens place. This gives <math>1 \cdot 5 \cdot 8 \cdot 7 = 280</math> possibilities. Together, the solution is <math>448 + 280 = \boxed{728}</math>.
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Case <math>1</math>: Thousands digit is even
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<math>4, 6</math>, two possibilities, then there are only <math>\frac{10}{2} - 1 = 4</math> possibilities for the units digit. This leaves <math>8</math> possible digits for the hundreds and <math>7</math> for the tens places, yielding a total of <math>2 \cdot 8 \cdot 7 \cdot 4 = 448</math>.
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Case <math>2</math>: Thousands digit is odd
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<math>5</math>, one possibility, then there are <math>5</math> choices for the units digit, with <math>8</math> digits for the hundreds and <math>7</math> for the tens place. This gives <math>1 \cdot 8 \cdot 7 \cdot 5= 280</math> possibilities.
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 +
Together, the solution is <math>448 + 280 = \boxed{728}</math>.
 +
== Solution 2 by PEKKA(more elaborate) ==
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Firstly, we notice that the thousands digit could be <math>4</math>, <math>5</math> or <math>6</math>.
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Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework.
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Case <math>1</math>:
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Here we let thousands digit be <math>4</math>.
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4 _ _ _
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We take care of restrictions first, and realize that there are 4 choices for the last digit, namely <math>2</math>,<math>6</math>,<math>8</math> and <math>0</math>.
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Now that there are no restrictions we proceed to find that there are <math>8</math> choices for the hundreds digit and <math>7</math> choices for the tens digit. Therefore we have <math> 8 \cdot 7 \cdot 4 = 224</math> numbers that satisfy the conditions posed by the problem.
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Case <math>2</math>
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Here we let thousands digit be <math>5</math>.
 +
 
 +
5 _ _ _
 +
 
 +
Again, we take care of restrictions first. This time there are 5 choices for the last digit, which are all the even numbers because 5 is odd.
 +
 
 +
Now that there are no restrictions we proceed to find that there are <math>8</math> choices for the hundreds digit and <math>7</math> choices for the tens digit. Therefore we have <math> 8 \cdot 7 \cdot 5 = 280</math> numbers that satisfy the conditions posed by the problem.
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 +
Case <math>3</math>
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 +
Here we let thousands digit be <math>6</math>.
 +
 
 +
6 _ _ _
 +
 
 +
Once again, we take care of restrictions first. Since 6 is even,  there are 4 choices for the last digit, namely <math>2</math>,<math>6</math>,<math>8</math> and <math>0</math>
 +
 
 +
Now we proceed to find that there are <math>8</math> choices for the hundreds digit and <math>7</math> choices for the tens digit. Therefore we have <math> 8 \cdot 7 \cdot 4 = 224</math> numbers that satisfy the conditions posed by the problem.
 +
 
 +
Now that we have our answers for each possible thousands place digit, we add up our answers and get <math>224+280+224</math>= <math>\boxed{728}</math>
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~PEKKA
  
 
== See also ==
 
== See also ==

Latest revision as of 20:20, 22 July 2021

Problem

How many even integers between 4000 and 7000 have four different digits?

Solution 1

The thousands digit is $\in \{4,5,6\}$.

Case $1$: Thousands digit is even

$4, 6$, two possibilities, then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 8 \cdot 7 \cdot 4 = 448$.


Case $2$: Thousands digit is odd

$5$, one possibility, then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 \cdot 8 \cdot 7 \cdot 5= 280$ possibilities.

Together, the solution is $448 + 280 = \boxed{728}$.

Solution 2 by PEKKA(more elaborate)

Firstly, we notice that the thousands digit could be $4$, $5$ or $6$. Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework. Case $1$:

Here we let thousands digit be $4$.

4 _ _ _

We take care of restrictions first, and realize that there are 4 choices for the last digit, namely $2$,$6$,$8$ and $0$. Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 4 = 224$ numbers that satisfy the conditions posed by the problem.

Case $2$

Here we let thousands digit be $5$.

5 _ _ _

Again, we take care of restrictions first. This time there are 5 choices for the last digit, which are all the even numbers because 5 is odd.

Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 5 = 280$ numbers that satisfy the conditions posed by the problem.

Case $3$

Here we let thousands digit be $6$.

6 _ _ _

Once again, we take care of restrictions first. Since 6 is even, there are 4 choices for the last digit, namely $2$,$6$,$8$ and $0$

Now we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 4 = 224$ numbers that satisfy the conditions posed by the problem.

Now that we have our answers for each possible thousands place digit, we add up our answers and get $224+280+224$= $\boxed{728}$ ~PEKKA

See also

1993 AIME (Problems • Answer Key • Resources)
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First question 		Followed by
Problem 2
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All AIME Problems and Solutions

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