Floor function

Revision as of 03:41, 29 August 2022 by Pikaniko (talk | contribs) (Intermediate Problems)

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

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)

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 \[\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right]+\ldots\]

(1986 IMO, #6)

See Also