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Math Help - Sum of the digits

  1. #1
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    Sum of the digits

    Prove that S(\overline{9\ldots 9})=S(\overline{9\ldots 9}^2) (the number of 9s are the same); S(n) is the sum of the digits of the number n.
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  2. #2
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    Hello, James!

    Prove that S(\overline{99\hdots 9})\:=\:S(\overline{99\hdots 9}^2) . . . the number of 9s is the same.

    S(N) is the sum of the digits of the number N.
    \text{Let }\,N \,=\,\underbrace{999\hdots 9}_{n\text{ digits}}

    . . Then: . \boxed{S(N) \,=\,9n}


    Since N \:=\:10^n-1, then: . N^2\;=\;(10^n-1)^2 \;=\;10^{2n} - 2\!\cdot\!10^n + 1

    . . That is: . N^2\:=\:10^n(10^n-2) + 1


    This is the number (10^n-2) "moved" n places to the left, plus a 1.

    \text{The number }10^n-2\text{ has the form: }\:\underbrace{999\hdots98}_{n\text{ digits}}
    . . It contains n-1 9's and an 8.


    Therefore: . n-1)9 + 8 + 1\quad\Rightarrow\quad\boxed{ S(N^2) \;=\;9n}" alt="S(N^2) \:=\n-1)9 + 8 + 1\quad\Rightarrow\quad\boxed{ S(N^2) \;=\;9n}" />

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