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

Latest revision as of 17:16, 28 March 2022

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 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 $6 \cdot 100 = \boxed{600}$ .

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 2 Shortcut

Because $2,4,6,8$ are all multiples of $2$ , we can speed things up. We only need to check up to $\frac{12}{24}$ , and the rest should repeat. As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{12}{24}$ . Similarly, this should repeat 100 times, for $\boxed{600}$

~N828335

Solution 2 Bigger Shortcut (Close to Solution 6)

We only need to check the numbers where it increments, namely $\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}$ . As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{1}{2}$ . Similarly, this should repeat 100 times, for $\boxed{600}$

~JeffersonJ

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

Let $x=\lfloor x\rfloor+\{x\}$ then $\begin{align*} \lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\ &=2\lfloor x\rfloor+4\lfloor x\rfloor+6\lfloor x\rfloor+8\lfloor x\rfloor+\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor\\ &=20\lfloor x\rfloor+(\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor) \end{align*}$ Similar to the previous solutions, the value of $\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor$ changes when $\{x\}=\frac{m}{n}$ , where $m\in\{1,2,3,...,n-1\}$ , $n\in\{2,4,6,8\}$ . Using Euler's Totient Function $\sum\limits_{k=0}^4 \phi(2k)$ to obtain $12$ different values for $\{x\}=\frac{m}{n}$ . (note that here Euler's Totient Function counts the number of $\{x\}=\frac{m}{n}$ where $m$ , $n$ are relatively prime so that the values of $\{x\}$ won't overlap.).

Thus if $k$ can be expressed as $\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor$ , then $k=20a+b$ for some non-negative integers $a$ , $b$ , where there are $12$ values for $b$ .

Exclusively, there are $49$ values for $a$ in the range $0<k<1000$ , or $49\cdot12=588$ ordered pairs $(a,b)$ .

If $a=0$ , $b\neq0$ , which includes $11$ ordered pairs.

If $a=50$ , $b=0$ , which includes $1$ ordered pair.

In total, there are $588+11+1=\boxed{600}$ values for $k$ .

~ Nafer

Solution 5

To simplify the question, let $y = 2x$ . Then, the expression in the question becomes $\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$ .

Let $\{x\}$ represent the non-integer part of $x$ (For example, $\{2.8\} = 0.8$ ). Then,

$\begin{align*} \lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor &= y - \{y\} + 2y - \{2y\} + 3y - \{3y\} + 4y - \{4y\} \\ &= 10y - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10(\lfloor y \rfloor + \{y\}) - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 10\{y\} - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) \\ \end{align*}$

Since $\lfloor y \rfloor$ is always an integer, $10\lfloor y \rfloor$ will be a multiple of 10. Thus, we look for the range of the other part of the expression. We will be able to reach the same numbers when $y$ ranges from $0$ to $1$ , because the curly brackets ( $\{\}$ ) gets rid of any integer part. Let the combined integer part of $2y$ , $3y$ , and $4y$ be $k$ (In other words, $k = \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$ ). Then,

$\begin{align*} 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) &= 9\{y\} - (2\{y\} + 3\{y\} + 4\{y\} - k) \\ &= 9\{y\} - (9\{y\} - k) \\ &= k \end{align*}$

The maximum value of $k$ will be when $y$ is slightly less than $1$ , which means $k = 1 + 2 + 3 = 6$ . As $y$ increases from $0$ to $1$ , $k$ will increase whenever $2y$ , $3y$ , or $4y$ is an integer, which happens when $y$ hits one of the numbers in the set $\left\{\dfrac14, \dfrac13, \dfrac12, \dfrac23, \dfrac34 \right\}$ . When $y$ reaches $\dfrac12$ , both $2y$ and $4y$ will be an integer, so $k$ will increase by $2$ . For all the other numbers in the set, $k$ increases by $1$ since only $1$ number in the set will be an integer. Thus, the possible values of $k$ are $\{0, 1, 2, 4, 5, 6\}$ .

Finally, looking back at the original expression, we plug in $k$ to get a multiple of $10$ plus any number in $\{0, 1, 2, 4, 5, 6\}$ . Thus, we hit all numbers ending in $\{0, 1, 2, 4, 5, 6\}$ , of which there are $\boxed{600}$ .

~Owen1204

Solution 6

Imagine that we gradually increase $x$ from $0$ to $1$ . At the beginning, the value of our expression is $0$ , at the end it is $2+4+6+8=20$ . Note that every time $x=\frac{a}{b}$ for some positive integer $a$ and a positive multiple $b$ of either $2, 4, 6,$ or $8$ . Thus, we have been able to express 12 of the integers from 1 through 20 when $0<x<1$ , namely when $x = \frac{1}{2}, \frac{2}{2}=1, \frac{1}{4}, \frac{3}{4}, \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \frac{5}{6}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}$ .

Notice that $\lfloor 2(x+n) \rfloor + \lfloor 4(x+n)\rfloor + \lfloor 6(x+n) \rfloor+ \lfloor8(x+n) \rfloor= \lfloor 2x \rfloor+ \lfloor 4x \rfloor+ \lfloor 6x \rfloor+ \lfloor 8x \rfloor+ 20n$ . This conceptually means that the number of integers that can be expressed in the range $(i, j)$ is the same as the number of integers that can be expressed in the range $(i+x, j+x)$ . Thus, because we can express $12$ integers in the range $(1, 20)$ , we can cover $12*50 = \boxed{600}$ from 1 to 1000. $-\text{thinker123}$

Solution 7

After observing, we can see that there are $6$ values of can be evaluated through the expression every $10$ numbers, so our answer is $6*100=600$ ~bluesoul

@@ Line 4: / Line 4: @@
 <math>\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor</math>,
-where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer less]] than or equal to <math>z</math>?
+where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest [[integer]] less than or equal to <math>z</math>?
-== Solution ==
+__TOC__
-{{solution}}
+== Solution 1 ==
+Noting that all of the numbers are even, we can reduce this to any real number <math>x</math> between <math>0</math> to <math>\frac 12</math>, as this will be equivalent to <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (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 <math>x = \frac 12</math> then we get the solution <math>10</math>; now consider when <math>x < \frac 12</math>: <math>\lfloor 2x \rfloor = 0</math>, <math>\lfloor 4x \rfloor \le 1</math>, <math>\lfloor 6x \rfloor \le 2</math>, <math>\lfloor 8x \rfloor \le 3</math>. But according to this the maximum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.
+*<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>.
+*<math>2</math>: Also simple, for example using <math>\frac 16</math>.
+*<math>3</math>: The partition must either be <math>1+1+1</math> or <math>1+2</math>. If <math>\lfloor 4x \rfloor = 1</math>, then <math>x \ge \frac 14</math>, but then <math>\lfloor 8x \rfloor \ge 2</math>; not possible; and vice versa to show that the latter partition doesn't work. So we cannot obtain <math>3</math>.
+*<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case we see that <math>\frac 14</math> works.
+*<math>5</math>: We can partition as <math>1+2+2</math>, from which we find that <math>\frac 13</math> works.
+*<math>6</math>: We can partition as <math>1+2+3</math>, from which we find that <math>\frac 38</math> works.
+Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers <math>1,2,4,5,6,10</math>; hence our solution is <math>6 \cdot 100 = \boxed{600}</math>.
+== Solution 2 ==
+As we change the value of <math>x</math>, the value of our [[expression]] changes only when <math>x</math> crosses [[rational number]] of the form <math>\frac{m}{n}</math>, where <math>n</math> is divisible by 2, 4, 6 or 8.  Thus, we need only see what happens at the numbers of the form <math>\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}</math>.  This gives us 24 calculations to make; we summarize the results here:
+<math>\frac{1}{24}, \frac{2}{24} \to 0</math>
+<math>\frac{3}{24} \to 1</math>
+<math>\frac{4}{24}, \frac{5}{24} \to 2</math>
+<math>\frac{6}{24}, \frac{7}{24} \to 4</math>
+<math>\frac{8}{24} \to 5</math>
+<math>\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6</math>
+<math>\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10</math>
+<math>\frac{15}{24} \to 11</math>
+<math>\frac{16}{24},\frac{17}{24} \to 12</math>
+<math>\frac{18}{24}, \frac{19}{24} \to 14</math>
+<math>\frac{20}{24}\to 15</math>
+<math>\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16</math>
+<math>\frac{24}{24} \to 20</math>
+Thus, we hit 12 of the first 20 integers and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.
+===Solution 2 Shortcut===
+Because <math>2,4,6,8</math> are all multiples of <math>2</math>, we can speed things up. We only need to check up to <math>\frac{12}{24}</math>, and the rest should repeat. As shown before, we hit 6 integers (<math>1,2,4,5,6,10</math>) from <math>\frac{1}{24}</math> to <math>\frac{12}{24}</math>. Similarly, this should repeat 100 times, for <math>\boxed{600}</math>
+~[[User:N828335|N828335]]
+===Solution 2 Bigger Shortcut (Close to Solution 6)===
+We only need to check the numbers where it increments, namely <math>\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}</math>. As shown before, we hit 6 integers (<math>1,2,4,5,6,10</math>) from <math>\frac{1}{24}</math> to <math>\frac{1}{2}</math>. Similarly, this should repeat 100 times, for <math>\boxed{600}</math>
+~JeffersonJ
+==Solution 3==
+Recall from Hermite's Identity that <math>\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor</math>. Then we can rewrite <math>\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</math>
+<math>+ \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</math>. There are <math>12</math> 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 <math>20</math>). Starting from every integer <math>x</math>, 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 <math>20</math> while only achieving <math>12</math> of those <math>20</math> values. We can conveniently shift the <math>1000</math> (since it can be achieved) to the position of the <math>0</math> so that there are only complete cycles of <math>20</math>, and the answer is <math>\frac {12}{20}\cdot1000 = \boxed{600}</math>.
+== Solution 4 ==
+Let <math>x=\lfloor x\rfloor+\{x\}</math> then
+<cmath>\begin{align*}
+\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\
+&=2\lfloor x\rfloor+4\lfloor x\rfloor+6\lfloor x\rfloor+8\lfloor x\rfloor+\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor\\
+&=20\lfloor x\rfloor+(\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor)
+\end{align*}</cmath>
+Similar to the previous solutions, the value of <math>\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor</math> changes when <math>\{x\}=\frac{m}{n}</math>, where <math>m\in\{1,2,3,...,n-1\}</math>, <math>n\in\{2,4,6,8\}</math>. Using [[Euler's Totient Function]]
+<cmath>\sum\limits_{k=0}^4 \phi(2k)</cmath>
+to obtain <math>12</math> different values for <math>\{x\}=\frac{m}{n}</math>. (note that here Euler's Totient Function counts the number of <math>\{x\}=\frac{m}{n}</math> where <math>m</math>, <math>n</math> are relatively prime so that the values of <math>\{x\}</math> won't overlap.).
+Thus if <math>k</math> can be expressed as <math>\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor</math>, then <math>k=20a+b</math> for some non-negative integers <math>a</math>, <math>b</math>, where there are <math>12</math> values for <math>b</math>.
+Exclusively, there are <math>49</math> values for <math>a</math> in the range <math>0<k<1000</math>, or <math>49\cdot12=588</math> ordered pairs <math>(a,b)</math>.
+If <math>a=0</math>, <math>b\neq0</math>, which includes <math>11</math> ordered pairs.
+If <math>a=50</math>, <math>b=0</math>, which includes <math>1</math> ordered pair.
+In total, there are <math>588+11+1=\boxed{600}</math> values for <math>k</math>.
+~ Nafer
+== Solution 5 ==
+To simplify the question, let <math>y = 2x</math>. Then, the expression in the question becomes <math>\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor</math>.
+Let <math>\{x\}</math> represent the non-integer part of <math>x</math> (For example, <math>\{2.8\} = 0.8</math>). Then,
+<cmath>\begin{align*}
+\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor &= y - \{y\} + 2y - \{2y\} + 3y - \{3y\} + 4y - \{4y\} \\
+&= 10y - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\
+&= 10(\lfloor y \rfloor + \{y\}) - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\
+&= 10\lfloor y \rfloor + 10\{y\} - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\
+&= 10\lfloor y \rfloor + 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) \\
+\end{align*}</cmath>
+Since <math>\lfloor y \rfloor</math> is always an integer, <math>10\lfloor y \rfloor</math> will be a multiple of 10. Thus, we look for the range of the other part of the expression. We will be able to reach the same numbers when <math>y</math> ranges from <math>0</math> to <math>1</math>, because the curly brackets (<math>\{\}</math>) gets rid of any integer part. Let the combined integer part of <math>2y</math>, <math>3y</math>, and <math>4y</math> be <math>k</math> (In other words, <math>k = \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor</math>). Then,
+<cmath>\begin{align*}
+\{y\} - (\{2y\} + \{3y\} + \{4y\}) &= 9\{y\} - (2\{y\} + 3\{y\} + 4\{y\} - k) \\
+&= 9\{y\} - (9\{y\} - k) \\
+&= k
+\end{align*}</cmath>
+The maximum value of <math>k</math> will be when <math>y</math> is slightly less than <math>1</math>, which means <math>k = 1 + 2 + 3 = 6</math>. As <math>y</math> increases from <math>0</math> to <math>1</math>, <math>k</math> will increase whenever <math>2y</math>, <math>3y</math>, or <math>4y</math> is an integer, which happens when <math>y</math> hits one of the numbers in the set <math>\left\{\dfrac14, \dfrac13, \dfrac12, \dfrac23, \dfrac34 \right\}</math>. When <math>y</math> reaches <math>\dfrac12</math>, both <math>2y</math> and <math>4y</math> will be an integer, so <math>k</math> will increase by <math>2</math>. For all the other numbers in the set, <math>k</math> increases by <math>1</math> since only <math>1</math> number in the set will be an integer. Thus, the possible values of <math>k</math> are <math>\{0, 1, 2, 4, 5, 6\}</math>.
+Finally, looking back at the original expression, we plug in <math>k</math> to get a multiple of <math>10</math> plus any number in <math>\{0, 1, 2, 4, 5, 6\}</math>. Thus, we hit all numbers ending in <math>\{0, 1, 2, 4, 5, 6\}</math>, of which there are <math>\boxed{600}</math>.
+~Owen1204
+==Solution 6==
+Imagine that we gradually increase <math>x</math> from <math>0</math> to <math>1</math>. At the beginning, the value of our expression is <math>0</math>, at the end it is <math>2+4+6+8=20</math>. Note that every time <math>x=\frac{a}{b}</math> for some positive integer <math>a</math> and a positive multiple <math>b</math> of either <math>2, 4, 6,</math> or <math>8</math>. Thus, we have been able to express 12 of the integers from 1 through 20 when <math>0<x<1</math>, namely when <cmath>x = \frac{1}{2}, \frac{2}{2}=1, \frac{1}{4}, \frac{3}{4}, \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \frac{5}{6}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}</cmath>.
+Notice that <cmath>\lfloor 2(x+n) \rfloor + \lfloor 4(x+n)\rfloor + \lfloor 6(x+n) \rfloor+ \lfloor8(x+n) \rfloor= \lfloor 2x \rfloor+ \lfloor 4x \rfloor+ \lfloor 6x \rfloor+ \lfloor 8x \rfloor+ 20n</cmath>. This conceptually means that the number of integers that can be expressed in the range <math>(i, j)</math> is the same as the number of integers that can be expressed in the range <math>(i+x, j+x)</math>. Thus, because we can express <math>12</math> integers in the range <math>(1, 20)</math>, we can cover <math>12*50 = \boxed{600}</math> from 1 to 1000.
+<math>-\text{thinker123}</math>
+==Solution 7==
+After observing, we can see that there are <math>6</math> values of can be evaluated through the expression every <math>10</math> numbers, so our answer is <math>6*100=600</math> ~bluesoul
 == See also ==
-* [[1985 AIME Problems/Problem 9 | Previous problem]]
+* [[Floor function]]
-* [[1985 AIME Problems/Problem 11 | Next problem]]
+{{AIME box|year=1985|num-b=9|num-a=11}}
-* [[1985 AIME Problems]]
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
+[[Category:Intermediate Number Theory Problems]]

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