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Revision as of 2:07:09, 21 November 2021 by 3.221.77.51
Revision as of 2:27:23, 21 November 2021 by Mdarrell
Transcribed second half of page
This revision marked as approved.
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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 accounted
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.