AIME #10 1985

[stupidityismygam]

How many of the first [unparseable or potentially dangerous latex formula] positive integers can be expressed in the form

[unparseable or potentially dangerous latex formula]

where [unparseable or potentially dangerous latex formula] is a real number, and [unparseable or potentially dangerous latex formula] denotes the greatest integer less than or equal to [unparseable or potentially dangerous latex formula]

[bradp]

java time.

[zefuri]

n00b time.

[Quelloquialism]

Nothing formal here...

First off--the value of the sum is monotonically increasing, so if a number is skipped as x increases, it's not ever going to appear for any value of x.

Second--we can see that the value of the sum only changes when any one of the floors change, which would be any time x (when expressed as a fraction) has a denominator equal to the coefficient of x inside the floor.

For the first floor: [unparseable or potentially dangerous latex formula]
For the second floor: [unparseable or potentially dangerous latex formula]
For the third floor: [unparseable or potentially dangerous latex formula]
For the fourth floor: [unparseable or potentially dangerous latex formula]

Now, because the sum is a monotonically increasing function of x, taking the union of all these sets and sorting the resulting set will give us the list of "useful" values for x--that is, when x changes in value.

[unparseable or potentially dangerous latex formula]

So, we iterate through the list, trying each value.

At x = 1/8, the fourth term increments. (1)
At x = 1/6, the third term increments. (2)
At x = 1/4, the second and fourth terms increment. (4)
At x = 1/3, the third term increments. (5)
At x = 3/8, the fourth term increments. (6)
At x = 1/2, all four terms increment. (10)

At this point, we realize that all of the floor values have simultaneously hit an integral value. This means that, from here, you can "factor out" the 10, and the cycle repeats itself. So, all solutions are equivalent to 0, 1, 2, 4, 5, or 6, modulo 10. Since this same cycle repeats every 10, 1000/10 = 100, and we have 6 solutions, the total number of solutions should be 6*100 = 600.

...is it 600? Please?

[bradp]

yeah...

you can get all the problems at http://www.artofproblemsolving.com/Foru ... s-182.html