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Alla [95]
3 years ago
9

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

Mathematics
1 answer:
Butoxors [25]3 years ago
5 0
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.
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