Question

Find all n∈IN such that the number of digits of n equal to n

Answer 1

Note that $\log_{10}1=0$, $\log_{10}10=1$, $\log_{10}100=2$, and so on. The function $\log_{10}x$ is continuous and increasing for all $x>0$, so when $1$, we have $0$; when $10$, $1$; and so on.

This means

$\lfloor\log_{10}x\rfloor=\begin{cases}0&\text{for }1\le x$

which means we can capture the number of digits of $n\in\mathbb N$ with the function $\lfloor\log_{10}n\rfloor+1$.

So the problem is the same as finding positive integer solutions to

$\lfloor\log_{10}n\rfloor+1=n$

We know that $n=1$ has one digit, so clearly this must be a solution. We need to show that this is the only solution.

Recall that $\dfrac{\mathrm d}{\mathrm dx}[\log_{10}x+1]=\dfrac1{\ln10\,x}$, while $\dfrac{\mathrm d}{\mathrm dx}[x]=1$. This means $\log_{10}x$ increases at a much slower rate than $x$ as $x\to\infty$. We know the two functions intersect when $x=1$. Therefore it's clear that $x>\log_{10}x+1$ for all $x>1$.

Now, it's always the case that $\lfloor f(x)\rfloor\le f(x)$, so we're essentially done:

$x>\log_{10}x+1\ge\lfloor\log_{10}x\rfloor+1$

which means there are no other solutions than $n=1$.