Armstrong number

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

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

\(\textcolor{green}{1}\textcolor{red}{6}^3 + \textcolor{orange}{5}0^3+ \textcolor{blue}{33}^3=\textcolor{green}{1}\textcolor{red}{6}\textcolor{orange}{5}0\textcolor{blue}{33}\)

\(\textcolor{green}{1}\textcolor{red}{66}^3 + \textcolor{orange}{5}00^3+ \textcolor{blue}{333}^3=\textcolor{green}{1}\textcolor{red}{66}\textcolor{orange}{5}00\textcolor{blue}{333}\)

\(\textcolor{green}{1}\textcolor{red}{666}^3 + \textcolor{orange}{5}000^3+ \textcolor{blue}{3333}^3=\textcolor{green}{1}\textcolor{red}{666}\textcolor{orange}{5}000\textcolor{blue}{3333}\)

and so on.

And here is the proof:

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

We have:

\(1...6^{3} \ =\left[ 10^{n} \ +\dfrac{2}{3}\left( 10^{n} -1\right)\right]^{3}\\ 5...0^{3} =5^{3} \times 10^{3n} =125\times 10^{3n}\\ 3...3^{3} =\left(\dfrac{10^{n+1} -1}{3}\right)^{3}\)

Then

\(A=\left[ 10^{n} \ +\dfrac{2}{3}\left( 10^{n} -1\right)\right]^{3} +125\times 10^{3n} +\left(\dfrac{10^{n+1} -1}{3}\right)^{3}\\ =\dfrac{500}{3} 10^{3n} -\dfrac{50}{3} 10^{2} n+\dfrac{10}{3} 10^{n} -\dfrac{1}{3}\)

By multiplying \(\dfrac {500} {3}\) with \(10^{3n}\) using decimal places, we get:

\(\dfrac{500}{3} 10^{3n} =166.\overline{6} \times 10^{3n} =166\ \underbrace{6...6}_{n\ 6's}\underbrace{6...6}_{n\ 6's}\underbrace{6...6}_{n\ 6's} .\overline{6}\\ \)

Likewise,

\(\dfrac{50}{3} 10^{2n} =16.\overline{6} \times 10^{2n} =16\underbrace{6...6}_{n\ 6's}\underbrace{6...6}_{n\ 6's} .\overline{6}\\ \)

\(\Rightarrow \dfrac{500}{3} 10^{3n} -\dfrac{50}{3} 10^{2n} =1\underbrace{6...6}_{n\ 6's} 5\underbrace{0...0}_{n\ 0's}\underbrace{0...0}_{n\ 0's}\\ \)

For the last term of the polynomial:

\(\dfrac{10}{3} 10^{n} -\dfrac{1}{3} =3.\overline{3} \times 10^{n} -0.\overline{3} =\underbrace{3...3}_{n+1\ 3's}\\ \)

Therefore,

\(A=\dfrac{500}{3} 10^{3n} -\dfrac{50}{3} 10^{2n} +\dfrac{10}{3} 10^{n} -\dfrac{1}{3} =1\underbrace{6...6}_{n\ 6's} 5\underbrace{0...0}_{n\ 0's}\underbrace{3...3}_{n+1\ 3's} \blacksquare \)

Note by Tín Phạm Nguyễn
3 weeks, 4 days ago

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This is very interesting

Abha Vishwakarma - 3 weeks, 1 day ago

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Where does the name "Armstrong Number" come from?

Blan Morrison - 3 weeks, 4 days ago

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I think the word "Armstrong" means repeated.

Ram Mohith - 3 weeks, 4 days ago

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

For example, \(153\) is an Armstrong number because \(153 = 1^3 + 5^3 + 3^3\)

Tín Phạm Nguyễn - 3 weeks, 4 days ago

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