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8-digit number and divisibility puzzle

  1. Jun 6, 2009 #1
    P is a 8-digit base ten positive integer having the form ABCDEFGH that uses each of the nonzero digits from 1 to 8 exactly once, and satisfies all of these conditions:

    (i) AB is divisible by 8.
    (ii) BC is divisible by 7.
    (iii) CD is divisible by 6.
    (iv) DE is divisible by 5.
    (v) EF is divisible by 4.
    (vi) FG is divisible by 3.
    (vii) GH is divisible by 2.

    Determine all possible value(s) that P can assume.
     
    Last edited: Jun 6, 2009
  2. jcsd
  3. Jun 6, 2009 #2

    Borek

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    Staff: Mentor

    72185634

    Force. Brute and dirty.
     
  4. Jun 6, 2009 #3
    Step 1 (list the multiplication table for XY < 90):
    AB has to be: 16,24,32,40,48,56,64,72,80,88
    BC has to be: 14,21,28,35,42,49,56,63,70,77,84,
    CD has to be: 12,18,24,30,36,42,48,54,60,66,72,78,84
    DE has to be: 10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85
    EF has to be: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88
    FG has to be: 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87
    GH has to be: 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88

    Step 2 Now eliminate any numbers that have 9 in them or 0 in them or numbers with the same digit twice:
    AB has to be: 16,24,32,48,56,64,72
    BC has to be: 14,21,28,35,42,56,63,84,
    CD has to be: 12,18,24,36,42,48,54,72,78,84
    DE has to be: 15,25,35,45,65,75,85
    EF has to be: 12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84
    FG has to be: 12, 15, 18, 21, 24, 27, 36, 42, 45, 48, 51, 54, 57, 63, 72, 75, 78, 81, 84, 87
    GH has to be: 12, 14, 16, 18, 24, 26, 28, 32, 34, 36, 38, 42, 46, 48, 52, 54, 56, 58, 62, 64, 68, 72, 74, 76, 78, 82, 84, 86

    Step 3:
    DE has only endings of 5, so E = 5
    Therefore EF is limited to 50 <= EF <= 59

    so EF has to be: 52, 56
    FG then has to be: 21, 24, 27, 63
    GH then has to be: 14, 16, 18, 32, 34, 38, 46, 48, 74, 76

    Step 4: Start making connections:

    If AB was 16, BC is 63, CD is 36, which can't be possible because of repeating
    If AB was 24, BC is 42, so nope
    If AB was 32, BC is 21 or 28, then CD is 12,18,or 84. 12 doesn't fit. So try 18 first. DE can then be 85 only. EF is 56, FG is 63. However 3 has shown up already, so try 84. DE is then 45, EF is then 56. Again FG is 63, so doesn't work.

    AB = 48, BC = 84, doesn't work
    AB = 56 => 63 => 36, doesn't work
    64 => 42 => 24 doesn't work.

    So AB = 72, so A = 7 B = 2
    BC = 21 or 28
    72 => 21 => 18 => 85 => 56 => 63 => 34

    So the final number is 72185634
     
  5. Aug 6, 2009 #4
    a=7 b=2 c=1 d=8 e=5 f=6 g=3 h=4
    I solved it by looking at the multiples of 8 for AB. Then the last digit of AB must equal the first digit of BC which is a multiple of 7. So I just checked the multiples and crossed out numbers that had either a zero or a digit that occured twice.
     
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