1993 AIME Problems/Problem 1

Problem

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

Solution

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 $448 + 280 = \boxed{728}$.

See also

1993 AIME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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