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Math Help - Senior Maths Challenge question

  1. #1
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    Senior Maths Challenge question

    Hello lads. So I did my senior maths challenge today, and there was a question which struck me as being difficult. I even asked my maths teacher and he couldn't solve it as fast as he could. So who can run me through how you would solve the following:

    abcd + abc + bcd + acd + abd + ab + bc + cd + ad + ac + bd+ a + b + c + d = 2009

    Find a+b+c+d

    Anyone got any idea how to do this?
    Last edited by mr fantastic; November 6th 2009 at 02:33 PM.
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  2. #2
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    Quote Originally Posted by 123ohrid View Post
    Hello lads. So I did my senior maths challenge today, and there was a question which struck me as being the sort of WTF questions. I even asked my maths teacher and he couldn't solve it as fast as he could. So who can run me through how you would solve the following:

    abcd + abc + bcd + acd + abd + ab + bc + cd + ad + ac + bd+ a + b + c + d = 2009

    Find a+b+c+d

    Anyone got any idea how to do this?
    Others may prove me wrong but I don't think there's a unique answer to this question. You'll notice that adding 1 to both sides gives

    (a+1)(b+1)(c+1)(d+1) = 2010.

    Some solutions

     <br />
a = 1, \;b = 1004, \; c = 0, \;d = 0,<br />

     <br />
a = 2, \;b = 669, \; c = 0, \;d = 0,<br />

     <br />
a = 4, \;b = 401, \; c = 0, \;d = 0,<br />

     <br />
a = 5, \;b = 334, \; c = 0, \;d = 0,<br />

     <br />
a = 9, \;b = 200, \; c = 0, \;d = 0.<br />

    We see that a+b+c+d changes in all these cases.
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  3. #3
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    I bet the question said none can equal 0. If so, 2010 breaks down into 4 prime factors which furthers this conclusion - 2,3,5,67.
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  4. #4
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    Quote Originally Posted by Jameson View Post
    I bet the question said none can equal 0. If so, 2010 breaks down into 4 prime factors which furthers this conclusion - 2,3,5,67.
    Nice observation!
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  5. #5
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    Yeah it sayd that a b c and d are positive integers.

    Also this might help, the answer is one of the following:

    A)73 B) 75 C) 77 D) 79 E) 81..

    So which one would it be?

    I'm guessing 77 because it's the sum of all your primes? But can you try to explain why?
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  6. #6
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    (a+1)(b+1)(c+1)(d+1) = 2010

    2010 has exactly 4 prime factors -- 2, 3, 5, 67 (this means that 2010 = 2\cdot 3 \cdot 5 \cdot 67. Convince yourself that you cannot write 2010 as another product of 4 different factors.), so (a+1) + (b+1) + (c+1) + (d+1) = 2 + 3 + 5 + 67 \Rightarrow a + b + c + d = 73
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