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 $\displaystyle \frac n2$ 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 $x < \frac 12$ : $\lfloor 2x \rfloor = 0$ , $\lfloor 4x \rfloor \le 1$ , $\lfloor 6x \rfloor \le 2$ , $\displaystyle \lfloor 8x \rfloor \le 3$ . 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 $x = \displaystyle \frac 18$ .
$2$ : Also simple, for example using $\displaystyle \frac 16$ .
$3$ : 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 $\frac{m}{\textrm{gcd}(2, 4, 6, 8)} = \frac{m}{24}$ . 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 $50 \cdot 12 = 600$ of the first 1000.

1985 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1985 AIME Problems/Problem 10

Problem

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Solution

Solution 1

Solution 2

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