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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 \(n\).

Mathematics
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For example: 9 = 0 + 9 40 = 36 + 4 + 0 120 = 117 + 2 + 0
I got you but 120 = 117 + 1 + 2 + 0
Oops.

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Other answers:

45 = 45 + 4 + 5 ???
No, 45 = 36 + 4 + 5
Kidding: 45 = 36 + 4 + 5
Do you use induction to prove?
what about the numbers below 9?
8 = 0 + 8
3 = 9(0) + 3
0 is a multiple of 9. 9 * 0 = 0
isnt that cheating lol? everything is a multiple of zero then
But I need to prove this thing. How do you do that?
98 = 90 + 9 + 8 ???
that's 107?
more like 81 + 9 + 8
98 = 81 + 9 + 8
How do you prove it?
\[\text{decimal} \]
Can you use Mathematical Induction to prove that? lol
I have no idea.
He is saying natural number @UnkleRhaukus
I don't think so Parth; you can't prove that with MI, not certain though.
I guess, Mukushla will give something to cheer upon..
i have a feeling it has something to do with mods
our number system is to base 10
sorry i lost my connection if n is a m+1 digit number \( n=(a_{m}a_{m-1}....a_{1}a_{0})=10^m a_{m}+10^{m-1} a_{m-1}+...+10a_{1}+a_{0} \) now \( n-(a_{m}+a_{m-1}+...+a_{1}+a_{0})=(10^m -1)a_{m}+(10^{m-1}-1)a_{m-1}+...+9a_{1}=9k \)
120=117+1+2+0
I 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.
Is my proof good enough?
10b*
@ParthKohli u did my work for m=2
\((10^k - 1)\) will be multiple of 9 always..
@waterineyes thats right
My guess is right: Mukushla has given something to cheer upon..
Yeah.

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