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Math Help - n = 13*(sum of digits)

  1. #1
    Senior Member DivideBy0's Avatar
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    n = 13*(sum of digits)

    The number of positive integers which are 13 times the sum of their digits is

    (A) 0
    (B) 1
    (C) 2
    (D) 3
    (E) 4

    This is some competition problem, AMC(2000). I know I should start by writing out the decimal expansions but I don't know how many digits the number has or what to do next.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    The number of positive integers which are 13 times the sum of their digits is

    (A) 0
    (B) 1
    (C) 2
    (D) 3
    (E) 4

    This is some competition problem, AMC(2000). I know I should start by writing out the decimal expansions but I don't know how many digits the number has or what to do next.
    Well it obviously isn't true for 1 digit. Try 2:
    "ab" = 10a + b

    So we want x, y such that
    10a + b  = 13(a + b)

    So
    b = -\frac{a}{4}
    which is ridiculous.

    So try 3 digits, abc:
    100a + 10b + c = 13(a + b + c)

    So
    c = {(100 - 13)a - (13 - 10)b}{12} = \frac{29a - b}{4}

    This, at least, might be possible. We need 29a - b to be divisible by 4 with a and b in the range from 0 - 9. By trial and error I immediately run into (a, b) = (1, 1):
    c = \frac{29 - 1}{4} = 7

    So 711 fits the bill.

    See if you can find others.

    What can you say about a 4 digit number?
    1000a + 100b + 10c + d = 13(a + b + c + d)

    Solve this for d and see if you can find any a, b, c for it and why/why not.

    -Dan
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  3. #3
    Senior Member DivideBy0's Avatar
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    Ok, that means...

    If N is a 4-digit number then

    1000a+100b+10c+d=13(a+b+c+d)

    \Rightarrow d = \frac{329a-29b-c}{4}

    Even if you have a = 1, b = 0, c = 9... d = 80. Any other values of a\neq 0, b, c only give larger values of d. I'm guessing this would only be exacerbated if N had > 4 digits.

    So N must have 3 digits

    Hohoho, I think there's a good way of doing this.

    As you said, c=\frac{29a-b}{4}

    Now a can only be 1, if it is any greater than we get the same trouble we had with 4 digit numbers, c will be too great.

    Then 29(1)-b \equiv 0 (mod 4), so 29 \equiv b (mod 4). That is, b\text{mod}4=1. So b=1,5,9

    Yes! So the solutions are 195, 117, 156. (D)
    Last edited by DivideBy0; September 21st 2007 at 07:03 AM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Ok, that means...

    If N is a 4-digit number then

    1000a+100b+10c+d=13(a+b+c+d)

    \Rightarrow d = \frac{329a-29b-c}{4}

    Even if you have a = 1, b = 0, c = 9... d = 80. Any other values of a\neq 0, b, c only give larger values of d. I'm guessing this would only be exacerbated if N had > 4 digits.

    So N must have 3 digits.

    Hohoho, I think there's a good way of doing this.

    As you said, c=\frac{29a-b}{4}

    Then 29a-b \equiv 0 (mod 4), so 29a \equiv b (mod 4). That is, b\text{mod}4=1. So b=1,5,9

    3 Solutions (D)!
    Bingo!

    -Dan
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  5. #5
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    Quote Originally Posted by DivideBy0 View Post
    This is some competition problem, AMC(2000). I know I should start by writing out the decimal expansions but I don't know how many digits the number has or what to do next.
    Which AMC? I presume this is AMC-10. You are in the 10th grade right?
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  6. #6
    Senior Member DivideBy0's Avatar
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    Yep, I'm in grade 10 and that's the Australian MC
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