The second Problem of the Week has now been posted. It's a harder version of last week's problem, but perhaps it is good for a bit of amusement. By its nature, it might be a bit hard to describe the solution in a comment, but have a look anyway.
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I found this one easier than the last one, because "region" allows you to simply count 20s. However I made the assumption that each roman symbol was its own number - so, for example, nine (IX) consists of an X and and I, and so for this exercise in counting to 20, it would count as 11 out of the requisite 80 'points.' Okay, now that that's out of the way...
First, draw a region around the ten and the X from the eleven (leaving out the one). X + X = 20.
Draw the second region to include the eight, nine, and the I from the eleven. X + V + I + I + I + I + I = 20.
Draw the third region to include the twelve, one, and seven.
The fourth region thus includes the two, three, four, five, and six.
I made the assumption that IX would count for nine, so must be split across two regions so that constituent figures could be counted separately to make the required total of 80. The most aesthetically pleasing solution I had under that constraint is basically the same as eric's, just with the roles of XI and IX reversed.
You fooled me on problem 1. I assumed you mean two lines representing clock HANDS (starting at the center). My answer was lines showing hands for the time 3:47. Oh well.
The URL for the answer to problem 2 has 'Spring15.' ;^}
Also, when I copy and paste the web site for problem 1's answer, Firefox says '404 page not found', and the URL has '%18rosenhjd' in it. Any idea why?
Did you really need to specify the sum? We know from the last time that the sum of the numbers on a click face is 78, but we need a multiple of 4. With Roman numbers there is no way to get the total down to 76 by splitting, but it's easy to get the extra 2 needed for 80 by splitting IX or IV. But whoops, IV is spelled IIII here...
Yes, this one is easy. Since the numbers sum to 78, you have to split the I and X in the 9 so that it sums to 11. This gives you the following regions:
1. The X from the 9 and the X from the 10 sum to 20.
2. The I from the 9 plus the 8 and 11 sum to 20.
3. The 1, 7, and 12 sum to 20.
4. The 2, 3, 4, 5, and 6 sum to 20.
Which is basically the same solution Zed gave.
Incidentally, the reason the 4 is usually spelled IIII in this context is to make the number of characters as symmetric as possible about the 12-6 axis. Moving away from the 12, you have 1-2-3-4-1 (totaling 11) on the right and 2-1-2-4-3 (totaling 12) on the left.