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 ?
Contents
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 .
Solution 3
Recall from Hermite's Identity that . Then we can rewrite . There are terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from ). Starting from every integer , we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by while only achieving of those values. We can conveniently shift the (since it can be achieved) to the position of the so that there are only complete cycles of , and the answer is .
Solution 4
Imagine that we increase from to . At the beginning, the value of our expression is , at the end it is . How many integers between and did we skip? We skip some integers precisely at those points where at least two of , , , and become integers at the same time.
Obviously, for and all four values become integers at the same time, hence we skip three integers at each of these locations. Additionally, for and the values and become integers at the same time, hence we skip one integer at each of the locations.
Therefore for we skip a total of integers. As in Solution 2, we conclude that we hit of the integers from to , and so we hit of the first .
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 |