Armstrong number

I just came across the post on the Facebook page Art of Mathematics, which says:

13+53+33=153\textcolor{#20A900}{1}^3 + \textcolor{#EC7300}{5}^3 + \textcolor{#3D99F6}{3}^3 = \textcolor{#20A900}{1}\textcolor{#EC7300}{5}\textcolor{#3D99F6}{3}

163+503+333=165033\textcolor{#20A900}{1}\textcolor{#D61F06}{6}^3 + \textcolor{#EC7300}{5}0^3+ \textcolor{#3D99F6}{33}^3=\textcolor{#20A900}{1}\textcolor{#D61F06}{6}\textcolor{#EC7300}{5}0\textcolor{#3D99F6}{33}

1663+5003+3333=166500333\textcolor{#20A900}{1}\textcolor{#D61F06}{66}^3 + \textcolor{#EC7300}{5}00^3+ \textcolor{#3D99F6}{333}^3=\textcolor{#20A900}{1}\textcolor{#D61F06}{66}\textcolor{#EC7300}{5}00\textcolor{#3D99F6}{333}

16663+50003+33333=166650003333\textcolor{#20A900}{1}\textcolor{#D61F06}{666}^3 + \textcolor{#EC7300}{5}000^3+ \textcolor{#3D99F6}{3333}^3=\textcolor{#20A900}{1}\textcolor{#D61F06}{666}\textcolor{#EC7300}{5}000\textcolor{#3D99F6}{3333}

and so on.

And here is the proof:

Let A=1...6n 6s3+5...0n 0s3+3...3n+1 3s3A=\underbrace{1...6}_{n\ 6's}^{3} +\underbrace{5...0}_{n\ 0's}^{3} +\underbrace{3...3}_{n+1 \ 3's}^{3}

We have:

1...63 =[10n +23(10n1)]35...03=53×103n=125×103n3...33=(10n+113)31...6^{3} \ =\left[ 10^{n} \ +\dfrac{2}{3}\left( 10^{n} -1\right)\right]^{3}\\ 5...0^{3} =5^{3} \times 10^{3n} =125\times 10^{3n}\\ 3...3^{3} =\left(\dfrac{10^{n+1} -1}{3}\right)^{3}

Then

A=[10n +23(10n1)]3+125×103n+(10n+113)3=5003103n503102n+10310n13A=\left[ 10^{n} \ +\dfrac{2}{3}\left( 10^{n} -1\right)\right]^{3} +125\times 10^{3n} +\left(\dfrac{10^{n+1} -1}{3}\right)^{3}\\ =\dfrac{500}{3} 10^{3n} -\dfrac{50}{3} 10^{2} n+\dfrac{10}{3} 10^{n} -\dfrac{1}{3}

By multiplying 5003\dfrac {500} {3} with 103n10^{3n} using decimal places, we get:

5003103n=166.6×103n=166 6...6n 6s6...6n 6s6...6n 6s.6\dfrac{500}{3} 10^{3n} =166.\overline{6} \times 10^{3n} =166\ \underbrace{6...6}_{n\ 6's}\underbrace{6...6}_{n\ 6's}\underbrace{6...6}_{n\ 6's} .\overline{6}\\

Likewise,

503102n=16.6×102n=166...6n 6s6...6n 6s.6\dfrac{50}{3} 10^{2n} =16.\overline{6} \times 10^{2n} =16\underbrace{6...6}_{n\ 6's}\underbrace{6...6}_{n\ 6's} .\overline{6}\\

5003103n503102n=16...6n 6s50...0n 0s0...0n 0s\Rightarrow \dfrac{500}{3} 10^{3n} -\dfrac{50}{3} 10^{2n} =1\underbrace{6...6}_{n\ 6's} 5\underbrace{0...0}_{n\ 0's}\underbrace{0...0}_{n\ 0's}\\

For the last term of the polynomial:

10310n13=3.3×10n0.3=3...3n+1 3s\dfrac{10}{3} 10^{n} -\dfrac{1}{3} =3.\overline{3} \times 10^{n} -0.\overline{3} =\underbrace{3...3}_{n+1\ 3's}\\

Therefore,

A=5003103n503102n+10310n13=16...6n 6s50...0n 0s3...3n+1 3sA=\dfrac{500}{3} 10^{3n} -\dfrac{50}{3} 10^{2n} +\dfrac{10}{3} 10^{n} -\dfrac{1}{3} =1\underbrace{6...6}_{n\ 6's} 5\underbrace{0...0}_{n\ 0's}\underbrace{3...3}_{n+1\ 3's} \blacksquare

Note by Tín Phạm Nguyễn
1 year, 5 months ago

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This is very interesting

Abha Vishwakarma - 1 year, 4 months ago

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The HADHA gene encodes the alpha subunit of the mitochondrial trifunctional protein (MTP), which is a lipid metabolism enzymes and thought to function in the mitochondria. However, HADHA was originally characterized as a gastrin-binding protein.

https://www.creative-biogene.com/genesearch/HADHA.html

Wendy Wilson - 5 months, 3 weeks ago

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Where does the name "Armstrong Number" come from?

Blan Morrison - 1 year, 5 months ago

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I think the word "Armstrong" means repeated.

Ram Mohith - 1 year, 5 months ago

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An Armstrong number is an integer of nn digits; the sum of its digits each one raised to the power of nn equals the integer itself.

For example, 153153 is an Armstrong number because 153=13+53+33153 = 1^3 + 5^3 + 3^3

Tín Phạm Nguyễn - 1 year, 5 months ago

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Bioanalysis is a term commonly used to describe the quantitative measurement of a drug or its metabolites in biological fluids. The determination of drug concentration in biological fluids provides data on the time course of drug action (PK) in animals and humans, which is an essential component of the drug discovery and development process.

https://www.creative-bioarray.com/dda-platform/bioanalytical-service.html

Bennie George - 4 months, 1 week ago

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