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

1993 AIME (Problems • Answer Key • Resources)
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1993 AIME Problems/Problem 1

Contents

Problem

Solution 1

Solution 2 by PEKKA(more elaborate)

See also