Copyright ©1995 by Paul Niquette. All rights reserved.
|even unknowns! The
previous sentence has an exclamation point; however, the
sophisticated solver will simply shrug, knowing that if
one is able to derive n equations based
on the conditions in the problem, one can merely solve
them 'simultaneously' to find the values of the n
unknowns. Large values of 'merely,' sometimes. And
not much fun. For seven unknowns, all one needs is
seven equations, and look what we have here...
That exclamation point is deserved, since the problem is possibly 'over-constrained' -- all bottled up, so to speak. But then without so much as one price given, a sophisticated solver might see this problem as 'under-constrained' -- a non-corker, so to speak. One might imagine prices spilling out all over the place.
Fortunately for us, there was a clue given in the statement of the puzzle -- that the lowest prices are to be found. Since one would not expect Sophisticated: The Magazine to publish a trivial puzzle, it is a minor supposition that the prices are neither equal to each other nor equal to zero. We may, however, wish to make an explicit assumption -- that the prices are integers.
f so, then we already know the price of one of the bottles: $1; however, we don't know which bottle goes for a buck. For the lowest possible integer solution, the rest of the bottles must be priced at $2, $3, $4, $5, $6, and $7, respectively -- 'respectively,' if we only knew which one gets what price. Now, there are seven possible prices for the bottle labeled A, six for B,... two for F, and finally one left over for bottle G. Mathematicians use an exclamation point for 'factorial' calculations -- and for good reason: 7! = 5,040. There must be a better way.
The puzzle designer favored sums in
pairs. Let's do the same. Each entry in the
bottom part of the table below makes explicit the
process for producing the respective sums shown as
entries in the top part of the table.
Lo and behold, we observe...
...only in numbers this time, not letters as given in the statement of the puzzle. Sophisticated solvers will notice four sums cannot be used in equations. In the table below, they are highlighted in bold.
the sum of 1 and 2 = 3 -- not the sum of two other
integers available to us. Similarly, the
sum of 1 and 3 = 4, which can be formed as the
sum of 2 and 2 all right -- but not the sum of two different
integers. The sum of 5 and 7 = 12, which
can be formed as the sum of 6 and 6, which are not
different from each other. Other formations for
12 require 8, 9, 10, or 11, which are all integers
that are not available to us. Likewise for the
sum of 6 and 7 = 13.
The same thing can be said of the
equations given in the puzzles: There are four missing
pairs of algebraic sums. They are easy to find, if you
remember your alphabet. Here they are alongside the
list of the missing numerical pairs.
Notice that D appears in two of the algebraic sums and G appears in the other two algebraic sums; likewise 1 appears in two of the numerical sums and 7 appears in the other two numerical sums. Whereas we do not know which letter corresponds with what number, we do know that D = 1 or 7 and G = 7 or 1. Not quite ready for a hoo-ha, though.
With two of the prices taken out of a
totally unkown category, here is a simpler table of
...and the remaining equations are
given in this table.
Similarly we are now able to simplify
the puzzle statement by removing for the time being
all equations in which the prices for D and G appear.
That leaves the following equations:
Again, we observe that four sums cannot
be used in equations -- neither the numerical
equations nor the algebraic equations...
...from which we deduce that A = 2 or 6
and F = 6 or 2. The problem has gotten a whole lot
smaller. With two more prices taken out of the
totally unkown category, that table of sums becomes...
There are no more opportunities to write equations. But here is where we are...
A + F = C + E and, since A + F = 8,...there is only one way that the prices of C and E can total $8 -- two ways, actually. Neither one will be able to use that $4 price. Accordingly, we now know that C = 3 or 5 and E = 5 or 3.
We know something else, too: That...
hat's a far cry
from the 5,040 combinations we started out with.
All that remains for us to do is to go back to the
original equations and figure out which combinations
in that table will satisfy them.
Here is the first equation...
A + B = D + E is satisfied if A = 2, F = 6, C = 3, E = 5, D = 1, and G = 7 -- oops!...which means we need to look at another equation. No big deal, we have twelve more...
A + C = B + D is satisfied if A = 2, F = 6, C = 3, E = 5, D = 1, and G = 7 -- oops!...and so it goes. Seems there are two solutions to this puzzle.
Don't you wonder what the prize is
for guessing the prices for those silly bottles?