Floor function

The greatest integer function, also known as the floor function, gives the greatest integer less than or equal to its argument. The floor of $x$ is usually denoted by $\lfloor x \rfloor$ or $[x]$ . The action of this function is the same as "rounding down." On a positive argument, this function is the same as "dropping everything after the decimal point," but this is not true for negative values.

Properties

$\lfloor a+b\rfloor\ge \lfloor a\rfloor+\lfloor b \rfloor$ for all real $(a,b)$ .
Hermite's Identity: $\lfloor na\rfloor = \left\lfloor a\right\rfloor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor$

Examples

$\lfloor 3.14 \rfloor = 3$

$\lfloor 5 \rfloor = 5$

$\lfloor -3.2 \rfloor = -4$

A useful way to use the floor function is to write $\lfloor x \rfloor=\lfloor y+k \rfloor$ , where y is an integer and k is the leftover stuff after the decimal point. This can greatly simplify many problems.

Alternate Definition

Another common definition of the floor function is

$\lfloor x \rfloor = x-\{x\}$

where $\{x\}$ is the fractional part of $x$ .

Problems

Introductory Problems

Let $[x]$ denote the largest integer not exceeding $x$ . For example, $[2.1]=2$ , $[4]=4$ and $[5.7]=5$ . How many positive integers $n$ satisfy the equation $\left[\frac{n}{5}\right]=\frac{n}{6}$ .

(2017 PCIMC)

Intermediate Problems

Find the integer $n$ satisfying $\left[\frac{n}{1!}\right]+\left[\frac{n}{2!}\right]+...+\left[\frac{n}{10!}\right]=1999$ . Here $[x]$ denotes the greatest integer less than or equal to $x$ .

(1999-2000 Hong Kong IMO Prelim)

What is the units (i.e., rightmost) digit of

$\left\lfloor \frac{10^{20000}}{10^{100}+3}\right\rfloor$ (1986 Putnam Exam, A-2)

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 $z$ ?

(1985 AIME)

Olympiad Problems

If $x$ is a positive real number, and $n$ is a positive integer, prove that

$[nx] \geq \frac{[x]}{1} + \frac{[2x]}{2} + \frac{[3x]}{3} + ... + \frac{[nx]}{n},$ where $[t]$ denotes the greatest integer less than or equal to $t$ .

(1981 USAMO, #5) (Discussion 1) (Discussion 2)

Let $[x]$ denote the integer part of $x$ , i.e., the greatest integer not exceeding $x$ . If $n$ is a positive integer, express as a simple function of $n$ the sum $\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right]+\ldots$

(1968 IMO, #6)

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