Notice this :

\(1\times9+2\)= \(11\)

\(12\times9+3\)= \(111\)

\(123\times9+4\)= \(1111\)

\(1234\times9+5\)= \(11111\)

\(12345\times9+6\)= \(111111\)

\(123456\times9+7\)= \(1111111\)

\(1234567\times9+8\)= \(11111111\)

\(12345678\times9+9\)= \(111111111\)

\(123456789\times9+10\)= \(1111111111\)

**Magic of mathematics**

Can anyone prove the cause of the sequence

Ps: without **without actual multiplication and Induction**

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## Comments

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TopNewestProof: Math is awesome. And we are done.

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LOLLOlLOLOLOLOLOL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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Hint: Mathematical Induction!

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Will you elaborate.

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Sure. I used induction on the number of digits in the number \(\overline{123.....n}\), which is \(1\) less than the number of ones, which, in turn, is the number with which we multiply \(\overline{123.....n}\) by, after adding \(9\). Firstly, I proved it true for the base case, that is, \(n=1\), then let it be true for \(1,2,3,...,k\), and then consecutively proved it right for \(n=k+1\). I've recently learnt induction, and am new to it, so I may have made a mistake. Please correct me if I have! Cheers :)

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@Satvik Golechha nice!! My teacher also told me but can anyone prove it without induction??

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seems like division algorithm

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Isn't it awesome!!

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question to which human brain can't answer #powerofmaths is 0/0 = ?

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Comment deleted Apr 12, 2015

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