Compare revisions: 1966_Sackson_352_November 28.jpg
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the rules of the latter and feel that the game is | the rules of the latter and feel that the game is | ||
played simply be keeping the positive and negative inform- | played simply be keeping the positive and negative inform- | ||
ation on the detective sheets. It seems almost auto- | ation on the detective sheets. It seems almost auto- | ||
matic. | matic. | ||
Claude called. He was at Arthur's. Nothing new. | Claude called. He was at Arthur's. Nothing new. | ||
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Solution of puzzle on 11/29 | |||
1) <u>A</u> has 2-SU and C has 1-SU, so all SU are accounted for. | |||
2) B has 2-UL which must be NUL & WUL since EUL & SUL are | |||
already accounted for. | |||
3) B has 1-UM but 0-EM. Since NUM and SUM are already ac- | |||
counted for, B must have WUM. | |||
4) B has 1-WR. B also has 3-W. Since B has already been shown | |||
to have WUL and WUM and the 3rd W is in the Rural area, | |||
B cannot have WIL or WIM. | |||
5) A has 1-NM. C has 3-N but 0-NF. Since NUL and NUM have | |||
already been accounted for, C's 3-N must be NIL, NRL, and | |||
either NUM or NRM. (It can be proven that C has NIM | |||
rather than NRM, but this not necessary for the solution.) | |||
6) C has 1-SI. This must be SIL since SIF and SIM are | |||
accounted for. | |||
7) C has 2-IL. Since C has already been shown to have NIL | |||
and SIL, C cannot have WIL or EIL. | |||
8) A, as stated before, has 2-SU. A also has 2-EI and 4-R. | |||
This limits the location of all 8 of A's cards to the SU, EI, | |||
and R areas. Among the cards that A cannot have is <u>WIL</u>, | |||
which has already been proven not to be in the possession | |||
of B or C. <u>WIL</u> is the solution. | |||
The statement, A had 2-W was thrown in as a small red | |||
herring. They must both be in the Rural area. |