Armstrong number

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

\(\textcolor{green}{1}^3 + \textcolor{orange}{5}^3 + \textcolor{blue}{3}^3 = \textcolor{green}{1}\textcolor{orange}{5}\textcolor{blue}{3}\)

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...6⏟n 6′sn+5...0⏟n 0′sn+3...3⏟n+1 3′snA=\underbrace{1...6}_{n\ 6's}^{n} +\underbrace{5...0}_{n\ 0's}^{n} +\underbrace{3...3}_{n+1 \ 3's}^{n}

We have:

1...63 =[10n +23(10n−1)]35...03=53×103n=125×103n3...33=(10n+1−13)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(10n−1)]3+125×103n+(10n+1−13)3=5003103n−503102n+10310n−13A=\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...6⏟n 6′s6...6⏟n 6′s6...6⏟n 6′s.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...6⏟n 6′s6...6⏟n 6′s.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}\\

⇒5003103n−503102n=16...6⏟n 6′s50...0⏟n 0′s0...0⏟n 0′s\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:

10310n−13=3.3‾×10n−0.3‾=3...3⏟n+1 3′s\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=5003103n−503102n+10310n−13=16...6⏟n 6′s50...0⏟n 0′s3...3⏟n+1 3′s■A=\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

#NumberTheory

Note by Tín Phạm Nguyễn
2 years, 8 months ago

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Comments

This is very interesting

Abha Vishwakarma - 2 years, 8 months ago

Where does the name "Armstrong Number" come from?

Blan Morrison - 2 years, 8 months ago

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I think the word "Armstrong" means repeated.

Ram Mohith - 2 years, 8 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 - 2 years, 8 months ago
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