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<x<10`, we have
`0<\log_(10)x<1`; when
`10<x<100`,
`1<\log_(10)x<2`; and so on.

This means


`\lfloor\log_(10)x\rfloor=\begin{cases}0&\text{for }1\le x<10\\1&\text{for }10\le x<100\\2&\text{for }100\le x<1000\\\vdots\end{cases}`

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
`(\mathrm d)/(\mathrm dx)[\log_(10)x+1]=\frac1{\ln10\,x}`, while
`(\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`.