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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
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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?
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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
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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
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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?
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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..
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