1993 AIME Problems/Problem 1

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 (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AIME Problems and Solutions

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