1985 AIME Problems/Problem 10
Problem
How many of the first 1000 positive integers can be expressed in the form
,
where is a real number, and denotes the greatest integer less than or equal to ?
Solution
We will be able to reach the same number of integers while ranges from 0 to 1 as we will when ranges from to for any integer (Quick proof: ). Since , the answer must be exactly 50 times the number of integers we will be able to reach as ranges from 0 to 1, including 1 but excluding 0.
Solution 1
Noting that all of the numbers are even, we can reduce this to any real number between to , as this will be equivalent to to for any integer (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10.
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):
We can match up the greatest integer functions with one of the partitions of the integer. If we let then we get the solution ; now consider when : , , , . But according to this the maximum we can get is , so we only need to try the first 6 numbers.
- : Easily possible, for example try plugging in .
- : Also simple, for example using .
- : The partition must either be or . If , then , but then ; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain .
- : We can partition as , and from the previous case we see that works.
- : We can partition as , from which we find that works.
- : We can partition as , from which we find that works.
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers ; hence our solution is .
Solution 2
As we change the value of , the value of our expression changes only when crosses rational number of the form , where is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form . This gives us 24 calculations to make; we summarize the results here:
Thus, we hit 12 of the first 20 integers and so we hit of the first 1000.
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |