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Math Help - numbertheory

  1. #1
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    numbertheory

    hi can anyone help me with this problem in number theory with regards to prime number? i don't know how to proive this..can someone help?

    Prove that an integer is divisible by 3 if and only if the sum of its digits is divbisble by 3. Prove that an integer is divisible by 9 if and only if the sum of its digits is divisible by 9...thanls
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  2. #2
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    Let N=[a_na_{n-1}...a_1a_0]_{10}. Then N=10^na_n+10^{n-1}a_{n-1}+...10a_1+a_0. Notice that 10 \equiv 1 \pmod 9 so 10^n \equiv 1 \pmod 9 and N=10^na_n+10^{n-1}+...10a_1+a_0 \equiv a_n+a_{n-1}+...+a_1+a_0 \pmod 9. That is, 9 \mid N iff 9 \mid a_n+a_{n-1}+...+a_1+a_0

    If you aren't familiar with congruences, notice that:
    N=10^na_n+10^{n-1}a_{n-1}+...10a_1+a_0 = \left  ((10^n-1)a_n+(10^{n-1}-1)a_{n-1}+..+9a_1 \right)+\left (a_n+a_{n-1}+...+a_1+a_0 \right )

    For divisibility by 3, notice that 10 \equiv 1 \pmod 3 and argue in the same way as in the case with divisibility by 9.
    Last edited by DavidEriksson; September 6th 2009 at 02:09 AM.
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  3. #3
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    Quote Originally Posted by DavidEriksson View Post
    Let N=[a_na_{n-1}...a_1a_0]_{10}. Then N=10^na_n+10^{n-1}a_{n-1}+...10a_1+a_0. Notice that 10 \equiv 1 \pmod 9 so 10^n \equiv 1 \pmod 9 and N=10^na_n+10^{n-1}+...10a_1+a_0 \equiv a_n+a_{n-1}+...+a_1+a_0 \pmod 9. That is, 9 \mid N iff 9 \mid a_n+a_{n-1}+...+a_1+a_0

    If you aren't familiar with congruences, notice that:
    N=10^na_n+10^{n-1}a_{n-1}+...10a_1+a_0 = \left  ((10^n-1)a_n+10^{n-1}a_{n-1}+..+9a_1 \right)+\left (a_n+a_{n-1}+...+a_1+a_0 \right )

    For divisibility by 3, notice that 10 \equiv 1 \pmod 3 and argue in the same way as in the case with divisibility by 9.

    thank you very much for the reply..Dave...really thanks
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