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Thread: 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 $\displaystyle a^2 + b^2 + c^2 = \frac{N}{11}$

    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...
    $\displaystyle 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 $\displaystyle abc $ is

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


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

    Let's check the first one :


    We have $\displaystyle a^2 + b^2 + c^2 = 9a+b $

    Since $\displaystyle c = b-a $ substitute it we obtain

    $\displaystyle a^2 + b^2 + (b-a)^2 = 9a + b $

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

    $\displaystyle (2b+9)^2 - 8(2b^2 - b) = k^2$

    $\displaystyle -12b^2 + 44b + 81 = k^2 $

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

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

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

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

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

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

    $\displaystyle 11^2 + 3(9)^2 $ and
    $\displaystyle 19^2 + 3(1)^2 $

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

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

    $\displaystyle c = 5-5 = 0$

    Therefore , the required number is $\displaystyle 550 $

    Use the similar method to see if there is another solution in the second case . It should be $\displaystyle 803 $ .
    Last edited by simplependulum; Jun 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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