Posers and Puzzles

Posers and Puzzles

  1. Joined
    03 Feb '05
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    59458
    01 Apr '09 13:52
    The sum of my 4 digit bank code equals the first two digits. The product equals the last two digits. What is my code?

    Solve same problem for a 5 digit code.
  2. Joined
    11 Nov '05
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    43938
    01 Apr '09 13:55
    Originally posted by camilli
    The sum of my 4 digit bank code equals the first two digits. The product equals the last two digits. What is my code?

    Solve same problem for a 5 digit code.
    You have to hold your bank code as a secret!
    Now everyone knows it and you have to change your code.
  3. Standard memberPalynka
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    01 Apr '09 14:45
    1236 => 1+2+3+6 = 12 and 1*2*3*6 = 36

    11612 => 1+1+6+1+2 = 11 and 1*1*6*1*2 = 12
    11324 => 1+1+3+2+4 = 11 and 1*1*3*2*4 = 24
    11315 => 1+1+3+1+5 = 11 and 1*1*3*1*5 = 15
  4. ALG
    Joined
    16 Dec '07
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    6190
    01 Apr '09 18:01
    Originally posted by Palynka
    1236 => 1+2+3+6 = 12 and 1*2*3*6 = 36

    11612 => 1+1+6+1+2 = 11 and 1*1*6*1*2 = 12
    11324 => 1+1+3+2+4 = 11 and 1*1*3*2*4 = 24
    11315 => 1+1+3+1+5 = 11 and 1*1*3*1*5 = 15
    How did you find it?
  5. Standard memberPalynka
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    02 Apr '09 10:126 edits
    Originally posted by Thomaster
    How did you find it?
    For the 5 digits - label them abcde:

    4*5*5=100, so even if two of the numbers are 1s, then the others must be lower than 4,5,5. Obviously reducing one of those and increasing the 1s is only making things worse as x^2 is convex. If that is so, then the sum of numbers must be lower than 1+1+4+5+5=16 => a=1.

    If a=1, then c+d+e=9 for all possible b!

    Moreover, b*c*d*e = 10*d+e <=> b*c = 10/e+1/d. b*c is an integer, so we need to find pairs of (e,d) that result in integers. These are {(1,2),(1,5),(2,4),(3,6)}. Let Sn be the set of solutions for that equation when b=n

    S1 = {(6,1,2),(3,1,5),(3,2,4),(2,3,6)}. But c+d+e=9, so we can eliminate (2,3,6). The other 3 are the solutions I've presented.

    For S2 to S9 is easy, we only need to divide the c in S1 by b, check if they are integers and if not, discard them. That leaves us with:

    S2 = {(3,1,2),(1,3,6)}. None sum up to 9. Eliminate.
    S3 = {(2,1,2),(1,2,4),(1,1,5)}. None sum up to 9. Eliminate.
    S4 = {}
    S5 = {}
    S6 = {(1,1,2)} Eliminate.
    Rest are empty sets.

    Three solutions then:
    11612
    11315
    11324

    Edit - Mmm... I assumed digits couldn't be zero. If they can, then the product is zero and d=e=0. We then have a+b+c=10*a+b <=> c=9*a as the only condition => a=1, c=9 and b can be anything. So we also have new possibilities:

    (1,b,9,0,0) for b ={0,1,...,9}
    (0,b,0,0,0) for b ={0,1,...,9}

    Edit - The statement about sum<16 is not correct. However, there are only a few combinations for which sum>=20 and product<100 and it's easy to check they don't work (e.g. 11199), so a=1.
  6. Standard memberPalynka
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    02 Apr '09 11:25
    For 6 digits there are no solutions. The same method can be applied, using the 4 valid integer combinations for (e,f) from 10/f + 1/e = some integer. a=1 for the same reasons as before.

    Then for b=1, go through the four (e,f) pairs and check possible (c,d) s.t c*d = (10*e+f)/(a*b*c*d) and c+d = (10*a+b)-a+b+e+f. Note that at each point, a,b,e,f are fixed.

    For example: a=1,b=1,e=1, f=2 => c+d = 6 and c*d=6 (note 11-(1+1+1+2)=6 and 12/(1*1*1*2)=6). No integers solve those two conditions.

    I leave the rest of the proof to you. 🙂
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