Tuesday, September 26, 2006

Clock Face

You have an analog clock whose face is number in a circle from 1 to 12, with 12 facing "north", 3 "east", 6 "south", and 9 "west". You are allowed to draw two lines which go all the way across the clock face, and divide the numbers on the clock into 3 or 4 groups depending on if the lines intersect. How can you draw the lines so that the numbers in each group add up to the same sum.


  1. One line just below the 11 and 2 making the first group 11, 12, 1, and 2 = 26

    Second line just above the 8 and 5. The second group, 3, 4, 9 & 10 = 25

    The bottom group, 5, 6, 7, & 8 also add up to 26.

  2. The sum of the numbers is 78. Since four doesn't go into 78 evenly, we can't have four groups. 78/3 is 26, so we need 3 groups of 26.

    Let's focus on the 12 first. if it is part of a consecutive group: 12+11+10 is too much. 11+12+1+2 works. 12+1+2+3+4+5 is too much. If the 12 is connected to an isolated group 12+2 +3+4+5 isolates the 1, no good. no other way to make 14. 12+1+6+7 isolates 2 - 5, too small, no good. 12+1+2 goes with 11, which we already have, or 5 and 6, which isolates 3,4. No good. 12+1+2+3 needs 10, which isolates 11. No good.

    Therefore one group must be 11+12+1+2.

    Next, focus on 10.
    10 + 9 + 8 too much.
    10 + 9 + 3 + 4. OK (this is a complete solution)

    10+3+4+5+6 is too big.


  3. mr don has the answer and Jonathan has a similar explanation to mine, so I'll leave it at that.

  4. What if we had a clock that showed 24 hours (in one big circle, with 1 and 13 opposite, not at the same spot). Can we divide with straight lines into n equal groups? For which n?

  5. If I am thinking correctly, you might have 12 groups (11 lines) that separate the groups into 1 and 24, 2 and 23, 3 and 22, 4 and 21, etc... so that each group adds up to 25.


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