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1993 AIME Problems/Problem 1

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Problem

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

Solution

The thousands digit is \in \{4,5,6\}. If the thousands digit is even (4,\ 6, 2 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 4 \cdot 8 \cdot 7 = 448.

If the thousands digit is odd (5, one possibility), then there is 5 choices for the units digit, with 8 digits for the hundreds and 7 for the tens place. This gives 1 \cdot 5 \cdot 8 \cdot 7 = 280 possibilities. Together, the solution is 448 + 280 = 728.

See also

1993 AIME (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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