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Mathematics 6 Online
Parth (parthkohli):

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 \(n\).

Parth (parthkohli):

For example: 9 = 0 + 9 40 = 36 + 4 + 0 120 = 117 + 2 + 0

OpenStudy (anonymous):

I got you but 120 = 117 + 1 + 2 + 0

Parth (parthkohli):

Oops.

OpenStudy (anonymous):

45 = 45 + 4 + 5 ???

Parth (parthkohli):

No, 45 = 36 + 4 + 5

OpenStudy (anonymous):

Kidding: 45 = 36 + 4 + 5

Parth (parthkohli):

Do you use induction to prove?

OpenStudy (lgbasallote):

what about the numbers below 9?

Parth (parthkohli):

8 = 0 + 8

OpenStudy (anonymous):

3 = 9(0) + 3

Parth (parthkohli):

0 is a multiple of 9. 9 * 0 = 0

OpenStudy (lgbasallote):

isnt that cheating lol? everything is a multiple of zero then

Parth (parthkohli):

But I need to prove this thing. How do you do that?

OpenStudy (anonymous):

98 = 90 + 9 + 8 ???

OpenStudy (lgbasallote):

that's 107?

OpenStudy (lgbasallote):

more like 81 + 9 + 8

OpenStudy (anonymous):

98 = 81 + 9 + 8

Parth (parthkohli):

How do you prove it?

OpenStudy (unklerhaukus):

\[\text{decimal} \]

OpenStudy (mimi_x3):

Can you use Mathematical Induction to prove that? lol

Parth (parthkohli):

I have no idea.

OpenStudy (anonymous):

He is saying natural number @UnkleRhaukus

OpenStudy (mimi_x3):

I don't think so Parth; you can't prove that with MI, not certain though.

OpenStudy (anonymous):

I guess, Mukushla will give something to cheer upon..

OpenStudy (lgbasallote):

i have a feeling it has something to do with mods

OpenStudy (unklerhaukus):

our number system is to base 10

OpenStudy (anonymous):

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 \( n-(a_{m}+a_{m-1}+...+a_{1}+a_{0})=(10^m -1)a_{m}+(10^{m-1}-1)a_{m-1}+...+9a_{1}=9k \)

OpenStudy (anonymous):

120=117+1+2+0

Parth (parthkohli):

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.

Parth (parthkohli):

Is my proof good enough?

Parth (parthkohli):

10b*

OpenStudy (anonymous):

@ParthKohli u did my work for m=2

OpenStudy (anonymous):

\((10^k - 1)\) will be multiple of 9 always..

OpenStudy (anonymous):

@waterineyes thats right

OpenStudy (anonymous):

My guess is right: Mukushla has given something to cheer upon..

Parth (parthkohli):

Yeah.

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