1. A: "No"
2. B: "No"
3. C: "No"
4. A: "No"
5. B: "Yes"
What are the colors of B's stamps?
A B C Statement rr rr gg 3 rr rg rg rr rg gg rr gg rr 2 rr gg rg 3 rr gg gg 1 rg rr rg 4 rg rr gg 4 rg rg rr rg rg rg rg rg gg rg gg rr 4 rg gg rg 4 gg rr rr 1 gg rr rg 3 gg rr gg 2 gg rg rr gg rg rg gg gg rr 3In the remaining seven combinations, B has one stamp of each color.
Eric Eilebrecht
provided this insightful solution, which I paraphrase:
The problem asks what colors B's stamps are. They must be one of
each. If they are both red, similar arguments could be used to
say they are both green. Since that sort of argument would lead
to an indeterminate color, B's stamps must be one of each.
[KD: I didn't include this as the final solution, but I have to admit it
uses the general puzzle assumption that "There must be a solution" to
arrive at the correct solution. The problem with it is I could change the
problem statement to read "B's stamps are RR, RG, or GG. What are they?",
and Eric's argument would again lead to RG as the solution.]
Rick Simineo provided the
following solution:
This takes quite a bit of explanation(at least the way I solved it):
There are 21 separate possible ways to affix the stamps.
(Three logicians can each have RR, RG, or GG=27 minus the six which
require more than four Red or Green stamps). Of these, the eight
which contain a matching set of doubles can be eliminated by this
logic:
If A saw B & C both wearing two Red Stamps, knowing that there were
only four Red Stamps to begin with, A should have easily deduced that
he was wearing two Green Stamps in the first round of questioning,
which he did not. The same holds true for inversion of color, and for
the other two logicians as well. This leaves 13 sets.
Of them, four of the ones containingopposite matches can be taken out
by this thought:
If A had GG and B had RR, then C should have thought this: 'If I have
two Red Stamps, then A easily knows what stamps he is wearing (see
above), yet he didn't. The same goes for two Green stamps with B.
I, therefore, must have one of each.' He did not say yes, so this must
not be the case. Similar arguments are made for other colors and for
A. (Please note that these two possibilities for B were not taken out,
B being the only one not to answer No after hearing both of the other
responses.) This leaves nine.
Of those nine, two of them have both A & C with one of either stamp,
and B with a pair. Had one of these situations been the case, B's
logic would have been this:
'If I have a pair, then A should have realized that he does not
have the same pair as me (first argument), since C said No. Also,
A must realize that he does not have a pair opposite mine (second
argument), also because C said no. Therefore, A should have known
that he has one of each, which he obviously didn't realize, having
said No. Therefore, I must not have a pair'
which eliminates those two situations, leaving us with seven.
"What of those seven?" you ask? Doesn't matter. All of them include B with a Stamp of each color, which is precisely the solution we were asked to find.