# A Number Theoretic Problem With the Floor Function

The floor function $\lfloor x\rfloor$ is defined for all real $x$ as an integer $n$ such that $n\le x\lt n+1.$ In other words, $\lfloor x\rfloor$ is the greatest integer not exceeding $x.$ $\lfloor x\rfloor =x$ only if $x$ is an integer; otherwise, $\lfloor x\rfloor\lt x.$

### Problem

$\displaystyle\bigg\lfloor\frac{n}{1}\bigg\rfloor +\bigg\lfloor\frac{n}{2}\bigg\rfloor +\bigg\lfloor\frac{n}{3}\bigg\rfloor +\ldots +\bigg\lfloor\frac{n}{n}\bigg\rfloor = t_{1}+t_{2}+\ldots +t_{n},$

where $t_{a}$ is the number of divisors of integer $a.$

### Solution 1

The proof is by induction. For $n=1$ the identity is obviously true. Assume it is true for $n=k:$

$\displaystyle\bigg\lfloor\frac{k}{1}\bigg\rfloor +\bigg\lfloor\frac{k}{2}\bigg\rfloor +\bigg\lfloor\frac{k}{3}\bigg\rfloor +\ldots +\bigg\lfloor\frac{k}{k}\bigg\rfloor = t_{1}+t_{2}+\ldots +t_{k},$

and let prove it for $n=k+1.$ For an integer $m,$ $0\gt m\le n+1,$ we need to consider two cases:

**$m$ divides $n+1$**$n+1 = ms,$ making $\displaystyle\bigg\lfloor\frac{n+1}{m}\bigg\rfloor =\frac{n+1}{m}=s$ whereas clearly $\displaystyle\bigg\lfloor\frac{n}{m}\bigg\rfloor =\frac{n}{m}=s-1,$ implying $\displaystyle\bigg\lfloor\frac{n+1}{m}\bigg\rfloor =\bigg\lfloor\frac{n}{m}\bigg\rfloor +1.$

**$m$ does not divide $n+1$**$n+1 = ms+r,$ $1\le r\lt n+1.$ Then $n = ms+(r-1),$ $1\le r\lt n+1.$ Therefore,

$\displaystyle\bigg\lfloor\frac{n+1}{m}\bigg\rfloor =s=\bigg\lfloor\frac{n}{m}\bigg\rfloor.$

In conclusion,

$\begin{cases} \displaystyle\bigg\lfloor\frac{n+1}{m}\bigg\rfloor =\bigg\lfloor\frac{n}{m}\bigg\rfloor + 1 & \mbox{if } m \mbox{ divides } n+1 ,\\ \displaystyle\bigg\lfloor\frac{n+1}{m}\bigg\rfloor =\bigg\lfloor\frac{n}{m}\bigg\rfloor & \mbox{if } m \mbox{ does not divide } n+1 . \end{cases}$

It follows that

$ \begin{align} \displaystyle\bigg\lfloor\frac{k+1}{1}\bigg\rfloor &+\bigg\lfloor\frac{k+1}{2}\bigg\rfloor +\bigg\lfloor\frac{k+1}{3}\bigg\rfloor +\ldots +\bigg\lfloor\frac{k+1}{k+1}\bigg\rfloor\\ &= \bigg\lfloor\frac{k}{1}\bigg\rfloor +\bigg\lfloor\frac{k}{2}\bigg\rfloor +\bigg\lfloor\frac{k}{3}\bigg\rfloor +\ldots +\bigg\lfloor\frac{k}{k}\bigg\rfloor +t_{k+1}, \end{align}$

which proves the statement.

### Solution 2

The number of terms in the sequence $1,2,3,\ldots,n$ divisible by a number $m$ equals $\displaystyle\bigg\lfloor\frac{n}{m}\bigg\rfloor.$ These are the numbers $m,2m,3m,\ldots$ Thus the expression

$\displaystyle\bigg\lfloor\frac{n}{1}\bigg\rfloor +\bigg\lfloor\frac{n}{2}\bigg\rfloor +\bigg\lfloor\frac{n}{3}\bigg\rfloor +\ldots +\bigg\lfloor\frac{n}{n}\bigg\rfloor$

sums up the total number of the integers not exceeding $n$ divisible by $1$ plus the number of those that are divisible by $2,$ and so on, effectively adding up the number of all the divisors of the members of the sequence $1,2,\ldots ,n$ i.e., $t_{1}+t_{2}+\ldots +t_{n}.$

### Solution 3

This proof is actually a geometric illustration for Solution #2.

Consider all the hyperbolas $y=k/x$ with integer $k\le n.$ The fact that point $(p,q)$ lies on the hyperbola $y=k/x$ means exactly that both $p$ and $q$ are among the divisors of $k.$ Thus the number of divisors of $k$ is the number of integer grid points on the graph of the hyperbola $y=k/x.$ Therefore the total number of the divisors of positive integers not exceeding $n$ is the number of grid points below and on the hyperbola $y=n/x.$

On the other hand, for a given $p,$ the number of points in the column $x=p$ below or on the hyperbola $y=n/x$ equals $\displaystyle\bigg\lfloor\frac{n}{p}\bigg\rfloor.$ Thus the statement at hand simply shows that the two ways of counting grid points is indeed the same.

### References

- D. O. Shklarsky, N. N. Chentzov, I. M. Yaglom, The USSR Olympiad Problem Book: Selected Problems and Theorems of Elementary Mathematics, Dover Publications; 3 Revised edition (September 28, 1993), Problem 103(a)

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