Armstrong number

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

$\textcolor{#20A900}{1}^3 + \textcolor{#EC7300}{5}^3 + \textcolor{#3D99F6}{3}^3 = \textcolor{#20A900}{1}\textcolor{#EC7300}{5}\textcolor{#3D99F6}{3}$

$\textcolor{#20A900}{1}\textcolor{#D61F06}{6}^3 + \textcolor{#EC7300}{5}0^3+ \textcolor{#3D99F6}{33}^3=\textcolor{#20A900}{1}\textcolor{#D61F06}{6}\textcolor{#EC7300}{5}0\textcolor{#3D99F6}{33}$

$\textcolor{#20A900}{1}\textcolor{#D61F06}{66}^3 + \textcolor{#EC7300}{5}00^3+ \textcolor{#3D99F6}{333}^3=\textcolor{#20A900}{1}\textcolor{#D61F06}{66}\textcolor{#EC7300}{5}00\textcolor{#3D99F6}{333}$

$\textcolor{#20A900}{1}\textcolor{#D61F06}{666}^3 + \textcolor{#EC7300}{5}000^3+ \textcolor{#3D99F6}{3333}^3=\textcolor{#20A900}{1}\textcolor{#D61F06}{666}\textcolor{#EC7300}{5}000\textcolor{#3D99F6}{3333}$

and so on.

And here is the proof:

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

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}\$/extract_itex] 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 1 year, 6 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote  # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or \[ ... $ to ensure proper formatting.
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This is very interesting

- 1 year, 6 months ago

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.

- 6 months, 4 weeks ago

Where does the name "Armstrong Number" come from?

- 1 year, 6 months ago

I think the word "Armstrong" means repeated.

- 1 year, 6 months ago

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 $153 = 1^3 + 5^3 + 3^3$

- 1 year, 6 months ago

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

- 5 months, 2 weeks ago