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Math Help - IMO 1960 problem 1

  1. #1
    Super Member Deadstar's Avatar
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    IMO 1960 problem 1

    Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of 11.

    Anyone know how to solve this one. We write N as abc then...

    Since N is divisible by 11 we know that a - b + c is divisible by 11. So a-b+c = 0 or 11.

    Then we also know that a^2 + b^2 + c^2 = \frac{N}{11}

    I can't get any sort of 'nice' solution from this and just end up with having to check 18 cases. (a+c=b, sub into above than a=0,1,2,... etc)
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Deadstar View Post
    ...and N/11 is equal to the sum of the squares of the digits of 11...
     1^2+1^2=2 ?

    I'm sure this is a typo.
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  3. #3
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    Maybe there is another method which is better ...

    the actual value of  abc is

     100a + 10b + c = (99a + 11b) + (a-b+c)


    which leads to two cases  a-b+c = 0 and  a-b+c = 11

    Let's check the first one :


    We have  a^2 + b^2 + c^2 = 9a+b

    Since  c = b-a substitute it we obtain

     a^2 + b^2 + (b-a)^2 = 9a + b

     2a^2 - a(2b+9) + 2b^2 - b = 0 which is a quadratic equation of a , check its discriminant and it should be a square  k^2 ie:

     (2b+9)^2 - 8(2b^2 - b) = k^2

     -12b^2 + 44b + 81 = k^2

     -(6b-11)^2 + 364 = 3k^2

     (6b-11)^2 + 3k^2 = 364 = 4 \times 91

    We have
     4 = 1^2 + 3(1)^2 = (1 + \sqrt{3} i )(1 - \sqrt{3} i )
     91 = 4^2 + 3(5)^2 = (4 + 5\sqrt{3} i )(4 - 5\sqrt{3} i )

    ( Note we neglect the case  4 = 2^2 + 3(0)^2 because later we will find that b must not be an integer . )

     (1 + \sqrt{3} i )( 4 + 5\sqrt{3} i ) = -11 + 9\sqrt{3}i
     (1 + \sqrt{3} i )( 4 + 5\sqrt{3} i ) = 19 - \sqrt{3} i

    Therefore , the two possible ways to write  364 in the form  x^2 + 3y^2 are :

     11^2 + 3(9)^2 and
     19^2 + 3(1)^2

    and we have  |6b-11| = 11 or  |6b-11| = 19 the only solution is  b=5 and  |k| = 1 ( corresponding to  b )

    so  a = \frac{2b+9 + 1}{4} or  \frac{2b+9 - 1}{4} which gives  a = 5

    c = 5-5 = 0

    Therefore , the required number is  550

    Use the similar method to see if there is another solution in the second case . It should be  803 .
    Last edited by simplependulum; June 7th 2010 at 06:57 PM.
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  4. #4
    Super Member Deadstar's Avatar
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    @ Chip yes should have been N.

    @ Simple pendulum nice solution! I don't have the answers so can't confirm but I never thought about introducing complex numbers...
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  5. #5
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    Quote Originally Posted by Deadstar View Post
    @ Chip yes should have been N.

    @ Simple pendulum nice solution! I don't have the answers so can't confirm but I never thought about introducing complex numbers...
    Human being never thought about introducing complex numbers but luckily we have Gauss !
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