1985 AIME Problems/Problem 10


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 $\boxed{600}$


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 $\boxed{600}$


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 $\boxed{600}$.


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

See also

1985 AIME (ProblemsAnswer KeyResources)
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