## 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.

Subscribe to:
Post Comments (Atom)

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

ReplyDeleteSecond 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.

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.

ReplyDeleteLet'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.

Done.

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

ReplyDeleteWhat 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?

ReplyDeleteIf 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.

ReplyDelete