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Math Help - using (mod)

  1. #1
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    using (mod)

    The sum of digits of the number 8215 is 8 + 2 +1 +5 = 7 (mod 9)
    Observe also 8215 = 9 9912) + 7= 7 (mod 9) does this hold for any number? Explain

    Suppose we want to compute the product 8215 x 3567 modulo 9. Replacing these numbers by those obtained by adding their digits and reducting modulo 9 gives 16 x 21 = 7 x 3 =21 = 3 (mod 9). Is it true that 8215 x 3567 = 3 (mod 9)? Explain
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    Quote Originally Posted by glost View Post
    The sum of digits of the number 8215 is 8 + 2 +1 +5 = 7 (mod 9)
    Observe also 8215 = 9*912 + 7= 7 (mod 9) does this hold for any number? Explain
    We know that for any positive integer a, 9 divides 10^a-1 (geometric sum)
    Therefore :

    10^a-1 \equiv 0 (\bmod 9)

    --> \boxed{10^a \equiv 1 (\bmod 9)}

    ---------------------
    Goin' back to the problem :

    8215=8\times \underbrace{10^3}_{\equiv 1}+2\times \underbrace{10^2}_{\equiv 1}+1\times \underbrace{10^1}_{\equiv 1}+5\times \underbrace{10^0}_{\equiv 1} \equiv \underbrace{8\times 1+2\times 1+1\times 1+5\times 1}_{8+2+1+5} (\bmod 9)

    Note that this is the sum of the digits of the number.

    8215 \equiv 16 (\bmod 9) \equiv \boxed{7} (\bmod 9)

    This holds for any number (try it out with a,b,c,d,... as the digits, you will see the same conclusion)


    Suppose we want to compute the product 8215 x 3567 modulo 9. Replacing these numbers by those obtained by adding their digits and reducting modulo 9 gives 16 x 21 = 7 x 3 =21 = 3 (mod 9). Is it true that 8215 x 3567 = 3 (mod 9)? Explain
    This uses basic rules of modular arithmetic (and the previous question) :

    a \equiv c (\bmod n) \implies ab \equiv bc (\bmod n)
    Last edited by Moo; June 30th 2008 at 11:12 AM.
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