# Expressing First 1000 Positive Integers as Floor Functions

• MHB
• anemone
In summary, the concept of expressing first 1000 positive integers as floor functions involves using the floor function to represent a large set of numbers using a single mathematical function. This concept is important in mathematics as it simplifies complicated expressions and can be used in various mathematical proofs and calculations. The formula for expressing the first 1000 positive integers as floor functions is ⌊x/1000⌋ + 1. This concept can also be applied to negative integers, where the result will be a negative integer. Some real-life applications of this concept include its use in computer science, engineering, and economics.
anemone
Gold Member
MHB
POTW Director
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?

Brute force! :LOL: :p

Python:
# -*- coding: utf-8 -*-

def f(x):
r = int(2*x)
r += int(4*x)
r += int(6*x)
r += int(8*x)
return r

def solve(n):
x0 = 0
f0 = -n
x1 = n
f1 = f(x1) - n
v = 0
while v < 201:
x = 0.5*(x0 + x1)
ff = f(x) - n
v += 1
if ff == 0:
return x
if ff < 0:
#root between x and x1
x0 = x
f0 = ff
if ff > 0:
#root between x0 and x
x1 = x
f1 = ff
return -1

sol = 0
nsol = 0
for i in range(1,1001):
t = solve(i)
if(t == -1):
nsol += 1
print(i, 'no solution')
else:
sol += 1
print(i, t)

print('')
print(sol, 'numbers possible')
print(nsol, 'numbers not possible')

Output as an attachment.

#### Attachments

• output.txt
16.9 KB · Views: 51
Let $f(x) = \lfloor 2x \rfloor+\lfloor 4x \rfloor+\lfloor 6x \rfloor+\lfloor 8x \rfloor$. It is an increasing function, taking only integer values. It has jumps at some multiples of $\frac1{24}$ and is constant everywhere else.

In the interval $0<x\leqslant\frac12$, $f(x)$ takes the value $0$ for $0<x<\frac3{24} = \frac18$. It then jumps to $1$ when $x=\frac3{24}$ (because $\lfloor 8x \rfloor$ becomes $1$ at that point). After that, it jumps to $2$ when $x = \frac4{24} = \frac16$. The next jump occurs at $x=\frac6{24}=\frac14$, when both $\lfloor 4x \rfloor$ and $\lfloor 8x \rfloor$ increase by $1$. So at that point $f(x)$ skips the value $3$ and goes straight from $2$ to $4$. There are then further jumps of $1$ at $x = \frac8{24}$ and $\frac9{24}$. After that, the next jump is at $x = \frac{12}{24} = \frac12$, when all four terms in $f(x)$ increase by $1$, and $f(x)$ jumps from $6$ to $10$, skipping over the values $7$, $8$ and $9$.

So in that interval from $0$ to $\frac12$, $f(x)$ increases by $10$, taking six integer values and skipping the other four. The same pattern then repeats in every interval of length $\frac12$: $f(x)$ will increase by $10$, taking six integer values and skipping the other four. So in the interval from $0$ to $50$, $f(x)$ will go from $0$ to $1000$, taking $600$ values and skipping $400$.

Thanks Opalg for your great insight and excellent solution!

Let $f(x)=\lfloor 2x \rfloor+\lfloor 4x \rfloor+\lfloor 6x \rfloor+\lfloor 8x \rfloor$--(1).

Observe that if $n$ is a positive integer, then from (1),

$f(x+n)=f(x)+20n$--(2) follows.

In particular, this means that if an integer $k$ can be expressed in the form $f(x_0)$ for some real $x_0$, then for $n=1,\,2,\,3,\cdots$ one can express $k+20n$ similarly, i.e. $k+20n=f(x_0)+20n=f(x_0+n)$.

In view of this, one may restrict attention to determining which of the first 20 positive integers are generated by $f(x)$ as $x$ ranges through the half-open interval $(0,\,1]$.

Next, observe that as $x$ increases, the value of $f(x)$ changes only when either $2,\,4x,\,6x$ or $8x$ attains an integral value, and that the change in $f(x)$ is always to a new, higher value. In the interval $(0,\,1]$, such changes occur precisely when $x$ is of the form $\dfrac{m}{n}$, where $1\le m \le n$ and $n=2,\,4,\,6,\,8$. There are 12 such fractions, in increasing order they are:

$\dfrac{1}{8},\,\dfrac{1}{6},\,\dfrac{1}{4},\,\dfrac{1}{3},\,\dfrac{3}{8},\,\dfrac{1}{2},\,\dfrac{5}{8},\,\dfrac{2}{3},\,\dfrac{3}{4},\,\dfrac{5}{6},\,\dfrac{7}{8}$ and $1$.

Therefore, only 12 of the first 20 positive integers can be represented in the desired form. Since $1000=50(20)$, in view of (2), this implies that in each of the 50 sequences,

$1,\,2,\,3,\,\cdots,20;21,\,22,\,23,\cdots,40;\cdots ;981,\,982,\,983,\,\cdots, 1000$

of 20 consecutive integers only 12 can be so expressed, leading to a total of $50(12)$ or 600 positive integers of the desired form.

## 1. What is a floor function?

A floor function, denoted by the symbol ⌊x⌋, is a mathematical function that rounds a number down to the nearest integer. It essentially gives the largest integer that is less than or equal to the given number.

## 2. How do you express the first 1000 positive integers as floor functions?

To express the first 1000 positive integers as floor functions, we use the formula ⌊x⌋ = x. So, the first 1000 positive integers can be written as ⌊1⌋, ⌊2⌋, ⌊3⌋, ..., ⌊1000⌋.

## 3. Why is it useful to express numbers as floor functions?

Expressing numbers as floor functions can be useful in various mathematical calculations and proofs. It allows us to simplify complex expressions and make them easier to work with. Floor functions are also commonly used in computer programming and data analysis.

## 4. Can floor functions only be used for positive integers?

No, floor functions can be used for any real numbers. However, when dealing with negative numbers, the result will be the next smallest integer. For example, ⌊-3.5⌋ = -4.

## 5. Are there any other types of rounding functions besides floor functions?

Yes, there are other types of rounding functions such as ceiling functions (⌈x⌉), which rounds a number up to the nearest integer, and rounding to the nearest integer function (⌊x + 0.5⌋), which rounds a number to the nearest integer. However, floor functions are the most commonly used type of rounding function.

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