1. Joined
    11 Nov '05
    Moves
    43938
    20 Jun '08 11:33
    You have a watch-face, numbered from 1 to 12 around a circle.

    Can you with two cuts divide the face so you get the numbers in groups with the same sum as the others?

    How do you do?
    Is there more solutions than one?

    Example: Cut in a cross and you get 1, 2, 3 in one group, 4, 5, 6 in another group etc. This is not a solution because 1+2+3=6 and 4+5+6=15 has not the same sum.
  2. Standard memberwolfgang59
    howling mad
    In the den
    Joined
    09 Jun '07
    Moves
    45641
    20 Jun '08 11:57
    Originally posted by FabianFnas
    You have a watch-face, numbered from 1 to 12 around a circle.

    Can you with two cuts divide the face so you get the numbers in groups with the same sum as the others?

    How do you do?
    Is there more solutions than one?

    Example: Cut in a cross and you get 1, 2, 3 in one group, 4, 5, 6 in another group etc. This is not a solution because 1+2+3=6 and 4+5+6=15 has not the same sum.
    Sum of all numbers is 78 which is not divisible by 4. We are therefore looking to divide the clockface into 3.

    78/3 = 26

    11,12,1,2

    9,10,3,4

    5,6,7,8

    Looks unique to me.
  3. Joined
    31 May '07
    Moves
    696
    20 Jun '08 15:21
    10,11,12,1,2,3 and 4,5,6,7,8,9

    you just use one cut to get half the way across.
  4. Joined
    15 Feb '07
    Moves
    667
    21 Jun '08 01:50
    Unfortunately, the problem is how to do it with exactly TWO cuts.

    I am wondering if there might be a second solution which involves cutting a two digit number into 2 separate numbers.. That might technically meet the qualifications.
  5. Joined
    31 May '07
    Moves
    696
    21 Jun '08 09:10
    it is two cuts. As I said, you use one cut to get half way across.
  6. Joined
    12 Sep '07
    Moves
    2668
    21 Jun '08 09:21
    Or have the two cuts at an angle to each other, not parallel.
  7. Joined
    05 Jun '07
    Moves
    906
    26 Jun '08 17:21
    I wonder if the stipulation requires the cuts to be a chord.
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