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ParthKohli
 4 years ago
I just discovered something which I am unable to prove.
Any natural number \(n\) can be represented as a multiple of 9 + Sum of the digits of \(n\).
ParthKohli
 4 years ago
I just discovered something which I am unable to prove. Any natural number \(n\) can be represented as a multiple of 9 + Sum of the digits of \(n\).

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ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0For example: 9 = 0 + 9 40 = 36 + 4 + 0 120 = 117 + 2 + 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got you but 120 = 117 + 1 + 2 + 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Kidding: 45 = 36 + 4 + 5

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Do you use induction to prove?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0what about the numbers below 9?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.00 is a multiple of 9. 9 * 0 = 0

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0isnt that cheating lol? everything is a multiple of zero then

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0But I need to prove this thing. How do you do that?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0more like 81 + 9 + 8

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0How do you prove it?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\text{decimal} \]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1Can you use Mathematical Induction to prove that? lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0He is saying natural number @UnkleRhaukus

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1I don't think so Parth; you can't prove that with MI, not certain though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess, Mukushla will give something to cheer upon..

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0i have a feeling it has something to do with mods

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0our number system is to base 10

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i lost my connection if n is a m+1 digit number \( n=(a_{m}a_{m1}....a_{1}a_{0})=10^m a_{m}+10^{m1} a_{m1}+...+10a_{1}+a_{0} \) now \( n(a_{m}+a_{m1}+...+a_{1}+a_{0})=(10^m 1)a_{m}+(10^{m1}1)a_{m1}+...+9a_{1}=9k \)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I got a proof, finally. \( \color{Black}{\Rightarrow \bar{abc} = 100a + 10a + c }\) \( \color{Black}{\Rightarrow 99a + a + 9b + b + c}\) \( \color{Black}{\Rightarrow 9(11a + b) + (a + b + c)}\) 11a + b is a multiple of 9.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Is my proof good enough?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli u did my work for m=2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\((10^k  1)\) will be multiple of 9 always..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@waterineyes thats right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My guess is right: Mukushla has given something to cheer upon..
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