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 \(n\).
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ParthKohli
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For example:
9 = 0 + 9
40 = 36 + 4 + 0
120 = 117 + 2 + 0
waterineyes
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I got you but 120 = 117 + 1 + 2 + 0
ParthKohli
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Oops.
waterineyes
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45 = 45 + 4 + 5 ???
ParthKohli
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No, 45 = 36 + 4 + 5
waterineyes
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Kidding:
45 = 36 + 4 + 5
ParthKohli
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Do you use induction to prove?
lgbasallote
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what about the numbers below 9?
ParthKohli
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8 = 0 + 8
waterineyes
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3 = 9(0) + 3
ParthKohli
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0 is a multiple of 9.
9 * 0 = 0
lgbasallote
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isnt that cheating lol? everything is a multiple of zero then
ParthKohli
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But I need to prove this thing. How do you do that?
waterineyes
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98 = 90 + 9 + 8 ???
lgbasallote
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that's 107?
lgbasallote
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more like 81 + 9 + 8
waterineyes
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98 = 81 + 9 + 8
ParthKohli
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How do you prove it?
UnkleRhaukus
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\[\text{decimal} \]
Mimi_x3
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Can you use Mathematical Induction to prove that? lol
ParthKohli
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I have no idea.
waterineyes
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He is saying natural number @UnkleRhaukus
Mimi_x3
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I don't think so Parth; you can't prove that with MI, not certain though.
waterineyes
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I guess, Mukushla will give something to cheer upon..
lgbasallote
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i have a feeling it has something to do with mods
UnkleRhaukus
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our number system is to base 10
mukushla
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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 \( n-(a_{m}+a_{m-1}+...+a_{1}+a_{0})=(10^m -1)a_{m}+(10^{m-1}-1)a_{m-1}+...+9a_{1}=9k \)
mukushla
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120=117+1+2+0
ParthKohli
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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.
ParthKohli
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Is my proof good enough?
ParthKohli
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10b*
mukushla
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@ParthKohli
u did my work for m=2
waterineyes
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\((10^k - 1)\) will be multiple of 9 always..
mukushla
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@waterineyes
thats right
waterineyes
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My guess is right:
Mukushla has given something to cheer upon..
ParthKohli
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Yeah.