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Math Help - Proving the theorem (if a numbers digits is divisible by 3, so is the number)

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    Proving the theorem (if a numbers digits is divisible by 3, so is the number)

    Use the theorem below to answer the question that follows.

    If the sum of a number's digits is divisible by three, the number is divisible by three.

    Which of the following ways of expressing a three digit number, n, best demonstrates why this theorem is true?

    A. n = a(10^2) + b(10) + c
    B. n = a(100) + b(10) + c
    C. n = a(99 + 1) + b(9 + 1) + c
    D. n = a(300-200) + b(30-20) + c(9-8)
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

    Quote Originally Posted by jbwtucker View Post
    Use the theorem below to answer the question that follows.

    If the sum of a number's digits is divisible by three, the number is divisible by three.

    Which of the following ways of expressing a three digit number, n, best demonstrates why this theorem is true?

    A. n = a(10^2) + b(10) + c
    B. n = a(100) + b(10) + c
    C. n = a(99 + 1) + b(9 + 1) + c
    D. n = a(300-200) + b(30-20) + c(9-8)
    C.

    Tell me why?
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    Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

    I'll do my best. I'm thinking...

    99 is divisible by 3. So if a(99 + 1) is divisible by 3, then a must be divisible by 3. Same goes for the b term.

    ...am I heading the right direction? Because c could be anything. So I don't see what that proves.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

    Quote Originally Posted by jbwtucker View Post
    I'll do my best. I'm thinking...

    99 is divisible by 3. So if a(99 + 1) is divisible by 3, then a must be divisible by 3. Same goes for the b term.

    ...am I heading the right direction? Because c could be anything. So I don't see what that proves.
    No.

    a(99+1)+b(9+1)+c

    =99a+a+9b+b+c

    =(99a+9b)+(a+b+c)

    =9(11a+b)+(a+b+c)

    Now, 9(11a+b) divisible by 3, so... What can you tell me...?
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    Re: Proving the theorem (if a numbers digits is divisible by 3, so is the number)

    I see. 9(11a+b) divided by 3 is 3(11a+b). And the original theorem states that the sum of a number's digits (a, b and c) is divisible by 3.

    Thanks!
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