1993 AIME Problems/Problem 1
Problem
How many even integers between 4000 and 7000 have four different digits?
Solution 1
The thousands digit is .
Case : Thousands digit is even
, two possibilities, then there are only possibilities for the units digit. This leaves possible digits for the hundreds and for the tens places, yielding a total of .
Case : Thousands digit is odd
, one possibility, then there are choices for the units digit, with digits for the hundreds and for the tens place. This gives possibilities.
Together, the solution is .
Solution 2 by PEKKA(more elaborate)
Firstly, we notice that the thousands digit could be , or . Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework. Case :
Here we let thousands digit be .
4 _ _ _
We take care of restrictions first, and realize that there are 4 choices for the last digit, namely ,, and . Now that there are no restrictions we proceed to find that there are choices for the hundreds digit and choices for the tens digit. Therefore we have numbers that satisfy the conditions posed by the problem.
Case
Here we let thousands digit be .
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 choices for the hundreds digit and choices for the tens digit. Therefore we have numbers that satisfy the conditions posed by the problem.
Case
Here we let thousands digit be .
6 _ _ _
Once again, we take care of restrictions first. Since 6 is even, there are 4 choices for the last digit, namely ,, and
Now we proceed to find that there are choices for the hundreds digit and choices for the tens digit. Therefore we have 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 = ~PEKKA
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
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