# Difference between revisions of "1985 AIME Problems/Problem 10"

## Problem

How many of the first 1000 positive integers can be expressed in the form

$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$,

where $x$ is a real number, and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$?

## Solution

We will be able to reach the same number of integers while $x$ ranges from 0 to 1 as we will when $x$ ranges from $n$ to $n + 1$ for any integer $n$ (Quick proof: $\lfloor 2(n+x)\rfloor + \ldots = \lfloor 2n + 2x\rfloor + \ldots = 2n + \lfloor 2x\rfloor \ldots$). Since $\lfloor 2\cdot50 \rfloor + \lfloor 4\cdot50 \rfloor + \lfloor 6\cdot50 \rfloor + \lfloor 8\cdot50 \rfloor = 100 + 200 + 300 + 400$, the answer must be exactly 50 times the number of integers we will be able to reach as $x$ 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 $x$ between $0$ to $\frac 12$, as this will be equivalent to $\frac n2$ to $\frac {n+1}2$ for any integer $n$ (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 $x = \frac 12$ then we get the solution $10$; now consider when $x < \frac 12$: $\lfloor 2x \rfloor = 0$, $\lfloor 4x \rfloor \le 1$, $\lfloor 6x \rfloor \le 2$, $\lfloor 8x \rfloor \le 3$. But according to this the maximum we can get is $1+2+3 = 6$, so we only need to try the first 6 numbers.

• $1$: Easily possible, for example try plugging in $x =\frac 18$.
• $2$: Also simple, for example using $\frac 16$.
• $3$: The partition must either be $1+1+1$ or $1+2$. If $\lfloor 4x \rfloor = 1$, then $x \ge \frac 14$, but then $\lfloor 8x \rfloor \ge 2$; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain $3$.
• $4$: We can partition as $1+1+2$, and from the previous case we see that $\frac 14$ works.
• $5$: We can partition as $1+2+2$, from which we find that $\frac 13$ works.
• $6$: We can partition as $1+2+3$, from which we find that $\frac 38$ works.

Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers $1,2,4,5,6,10$; hence our solution is $6 \cdot 100 = 600$.

### Solution 2

As we change the value of $x$, the value of our expression changes only when $x$ crosses rational number of the form $\frac{m}{n}$, where $n$ is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form $\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}$. This gives us 24 calculations to make; we summarize the results here:

$\frac{1}{24}, \frac{2}{24} \to 0$

$\frac{3}{24} \to 1$

$\frac{4}{24}, \frac{5}{24} \to 2$

$\frac{6}{24}, \frac{7}{24} \to 4$

$\frac{8}{24} \to 5$

$\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6$

$\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10$

$\frac{15}{24} \to 11$

$\frac{16}{24},\frac{17}{24} \to 12$

$\frac{18}{24}, \frac{19}{24} \to 14$

$\frac{20}{24}\to 15$

$\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16$

$\frac{24}{24} \to 20$

Thus, we hit 12 of the first 20 integers and so we hit $50 \cdot 12 = \boxed{600}$ of the first $1000$.

### Solution 3

Recall from Hermite's Identity that $\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor$. Then we can rewrite $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor$ $+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor$. There are $12$ 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 $20$). Starting from every integer $x$, 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 $20$ while only achieving $12$ of those $20$ values. We can conveniently shift the $1000$ (since it can be achieved) to the position of the $0$ so that there are only complete cycles of $20$, and the answer is $\frac {12}{20}\cdot1000 = \boxed{600}$.

### Solution 4

Imagine that we increase $x$ from $0$ to $1$. At the beginning, the value of our expression is $0$, at the end it is $2+4+6+8=20$. How many integers between $1$ and $20$ did we skip? We skip some integers precisely at those points where at least two of $2x$, $4x$, $6x$, and $8x$ become integers at the same time.

Obviously, for $x=1/2$ and $x=1$ all four values become integers at the same time, hence we skip three integers at each of these locations. Additionally, for $x=1/4$ and $x=3/4$ the values $4x$ and $8x$ become integers at the same time, hence we skip one integer at each of the locations.

Therefore for $x\in(0,1]$ we skip a total of $3+3+1+1=8$ integers. As in Solution 2, we conclude that we hit $12$ of the integers from $1$ to $20$, and so we hit $50 \cdot 12 = 600$ of the first $1000$.