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**

## Comments

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TopNewestProof: Math is awesome. And we are done. – Marc Vince Casimiro · 2 years, 1 month ago

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– Parth Lohomi · 2 years, 1 month ago

LOLLOlLOLOLOLOLOL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Log in to reply

Hint: Mathematical Induction! – Satvik Golechha · 2 years ago

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– Siddharth Bhatt · 2 years ago

Will you elaborate.Log in to reply

– Satvik Golechha · 2 years ago

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 :)Log in to reply

@Satvik Golechha nice!! My teacher also told me but can anyone prove it without induction?? – Parth Lohomi · 2 years ago

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seems like division algorithm – Nihar Mahajan · 2 years, 1 month ago

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– Parth Lohomi · 2 years, 1 month ago

Isn't it awesome!!Log in to reply

– Nihar Mahajan · 2 years, 1 month ago

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

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– Nihar Mahajan · 2 years, 1 month ago

well! then all questions in this world can be answered in this format .. :PLog in to reply

– Jai Gupta · 2 years, 1 month ago

Lololololol!!Log in to reply

– Siddharth Bhatt · 2 years, 1 month ago

Ye sure!!Log in to reply