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ParthKohli

  • 3 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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  1. ParthKohli
    • 3 years ago
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    For example: 9 = 0 + 9 40 = 36 + 4 + 0 120 = 117 + 2 + 0

  2. waterineyes
    • 3 years ago
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    I got you but 120 = 117 + 1 + 2 + 0

  3. ParthKohli
    • 3 years ago
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    Oops.

  4. waterineyes
    • 3 years ago
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    45 = 45 + 4 + 5 ???

  5. ParthKohli
    • 3 years ago
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    No, 45 = 36 + 4 + 5

  6. waterineyes
    • 3 years ago
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    Kidding: 45 = 36 + 4 + 5

  7. ParthKohli
    • 3 years ago
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    Do you use induction to prove?

  8. lgbasallote
    • 3 years ago
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    what about the numbers below 9?

  9. ParthKohli
    • 3 years ago
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    8 = 0 + 8

  10. waterineyes
    • 3 years ago
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    3 = 9(0) + 3

  11. ParthKohli
    • 3 years ago
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    0 is a multiple of 9. 9 * 0 = 0

  12. lgbasallote
    • 3 years ago
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    isnt that cheating lol? everything is a multiple of zero then

  13. ParthKohli
    • 3 years ago
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    But I need to prove this thing. How do you do that?

  14. waterineyes
    • 3 years ago
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    98 = 90 + 9 + 8 ???

  15. lgbasallote
    • 3 years ago
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    that's 107?

  16. lgbasallote
    • 3 years ago
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    more like 81 + 9 + 8

  17. waterineyes
    • 3 years ago
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    98 = 81 + 9 + 8

  18. ParthKohli
    • 3 years ago
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    How do you prove it?

  19. UnkleRhaukus
    • 3 years ago
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    \[\text{decimal} \]

  20. Mimi_x3
    • 3 years ago
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    Can you use Mathematical Induction to prove that? lol

  21. ParthKohli
    • 3 years ago
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    I have no idea.

  22. waterineyes
    • 3 years ago
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    He is saying natural number @UnkleRhaukus

  23. Mimi_x3
    • 3 years ago
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    I don't think so Parth; you can't prove that with MI, not certain though.

  24. waterineyes
    • 3 years ago
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    I guess, Mukushla will give something to cheer upon..

  25. lgbasallote
    • 3 years ago
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    i have a feeling it has something to do with mods

  26. UnkleRhaukus
    • 3 years ago
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    our number system is to base 10

  27. mukushla
    • 3 years ago
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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 \( n-(a_{m}+a_{m-1}+...+a_{1}+a_{0})=(10^m -1)a_{m}+(10^{m-1}-1)a_{m-1}+...+9a_{1}=9k \)

  28. mukushla
    • 3 years ago
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    120=117+1+2+0

  29. ParthKohli
    • 3 years ago
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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.

  30. ParthKohli
    • 3 years ago
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    Is my proof good enough?

  31. ParthKohli
    • 3 years ago
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    10b*

  32. mukushla
    • 3 years ago
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    @ParthKohli u did my work for m=2

  33. waterineyes
    • 3 years ago
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    \((10^k - 1)\) will be multiple of 9 always..

  34. mukushla
    • 3 years ago
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    @waterineyes thats right

  35. waterineyes
    • 3 years ago
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    My guess is right: Mukushla has given something to cheer upon..

  36. ParthKohli
    • 3 years ago
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    Yeah.

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