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Question Number 174928 by Mastermind last updated on 14/Aug/22

$$\mathrm{How}\mathrm{many}\mathrm{digits}\mathrm{does}{1000}^{1000}\mathrm{have}?$$$$$$$$$$$$\mathrm{Mastermind}$$

Answered by Ar Brandon last updated on 14/Aug/22

$$3001$$

Commented by Mastermind last updated on 14/Aug/22

$$\mathrm{your}\mathrm{solution}\mathrm{pls}$$

Answered by BaliramKumar last updated on 14/Aug/22

$${10}^1=10\left(2digits\right)$$$${10}^2=100\left(3digits\right)$$$${10}^3=1000\left(4digits\right)$$$$.....................$$$${1000}^{1000}={\left({10}^3\right)}^{1000}={10}^{3000}\left(3001digits\right)$$

Commented by Frix last updated on 14/Aug/22

$$\log {1000}^{1000}=1000\log 1000=3000$$$$\mathrm{we}\mathrm{have}\mathrm{a}‘‘1^{″}\mathrm{and}3000‘‘0^{″}\mathrm{s}\Rightarrow 3001\mathrm{digits}$$$$\mathrm{easy}.$$$$\mathrm{but}\mathrm{how}\mathrm{many}\mathrm{digits}\mathrm{does}{n}^n\mathrm{with}n\in \mathbb{N}\mathrm{have}?$$

Commented by mr W last updated on 14/Aug/22

$$n^{n}has\lfloor n\log n\rfloor +1digits$$

Commented by Ari last updated on 14/Aug/22

$$3^{3}has3ln3+1digits...oh..no$$

Commented by mr W last updated on 14/Aug/22

$$typo!$$$$\lfloor n\log n\rfloor +1$$

Commented by Ari last updated on 14/Aug/22

$$3^3=27,hastwodigits$$$$3log3+1\ne 2$$

Commented by Ari last updated on 14/Aug/22

$$perhapsrounded!$$

Commented by Ari last updated on 14/Aug/22

$$takethegraatestintegerof\left[nlogn\right]+1$$

Commented by mr W last updated on 15/Aug/22

$$pleasetrytounderstanditbefore$$$$yousaysomethingiswrong!$$$$iwroteclearly\lfloor n\log n\rfloor +1.$$

Commented by Ari last updated on 15/Aug/22

$$Ok!$$

Answered by princeDera last updated on 14/Aug/22

$$$$$$\lfloor 1000\log \left(1000\right)\rfloor +1=1000\lfloor 3\log \left(10\right)\rfloor +1=3000+1=3001$$

Commented by Ari last updated on 14/Aug/22

$$wrong!$$

Commented by princeDera last updated on 14/Aug/22

$$itwasamistake$$$$$$$${\log }_{10}andnot\log {}_e$$

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