Bottle Royal

Copyright 1995 by Paul Niquette. All rights reserved.


 
Seven 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...
 
Thirteen
A + E = B + C
B + C = D + F
Equations!
A + F = C + E
B + E = C + F
A + B = D + E
A + F = D + G
B + F = C + G
A + C = B + D
A + G = B + E
B + G = E + F
A + E = D + F
A + G = C + F
C + E = D + G

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.
 


$1
$2
$3
$4
$5
$6
$7
$1

3
4
5
6
7
8
$2
1+2

5
6
7
8
9
$3
1+3
2+3

7
8
9
10
$4
1+4
2+4
3+4

9
10
11
$5
1+5
2+5
3+5
4+5

11
12
$6
1+6
2+6
3+6
4+6
5+6

13
$7
1+7
2+7
3+6
4+7
5+7
6+7

 Lo and behold, we observe...
 

Thirteen
1 + 6 = 3 + 4
2 + 7 = 3 + 6
Equations!
1 + 7 = 2 + 6
2 + 7 = 4 + 5
1 + 4 = 2 + 3
1 + 7 = 3 + 5
3 + 6 = 4 + 5
1 + 5 = 2 + 4
2 + 5 = 3 + 4
3 + 7 = 4 + 6
1 + 6 = 2 + 5
2 + 6 = 3 + 5
4 + 7 = 5 + 6

...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.

pecifically, 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.
 


$1
$2
$3
$4
$5
$6
$7
$1

3
4
5
6
7
8
$2
1+2

5
6
7
8
9
$3
1+3
2+3

7
8
9
10
$4
1+4
2+4
3+4

9
10
11
$5
1+5
2+5
3+5
4+5

11
12
$6
1+6
2+6
3+6
4+6
5+6

13
$7
1+7
2+7
3+6
4+7
5+7
6+7

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.
 

A + D  1 + 2
C + D 1 + 3
E + G 5 + 7
F + G 6 + 7

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 sums...
 


$2
$3
$4
$5
$6
$2

5
6
7
8
$3
2+3

7
8
9
$4
2+4
3+4

9
10
$5
2+5
3+5
4+5

11
$6
2+6
3+6
4+6
5+6

...and the remaining equations are given in this table.
 

Only Three
Equations!
2 + 5 = 3 + 4
2 + 6 = 3 + 5
3 + 6 = 4 + 5

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:
 

Only Three
Equations!
A + E= B + C
A + F = C + E
B + E = C + F

Again, we observe that four sums cannot be used in equations -- neither the numerical equations nor the algebraic equations...
 

A + B 2 + 3
A + C 2 + 4
B + F 4 + 6
E + F 5 + 6

...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...
 


$3
$4
$5
$3

7
8
$4
3+4

9
$5
3+5
4+5

There are no more opportunities to write equations. But here is where we are...

  • D = 1 or 7 and G = 7 or 1; therefore D + G = 8.
  • A = 2 or 6 and F = 6 or 2; therefore A + F = 8.
  • There are four prices taken out of the totally unknown category: $1, $2, $6, and $7.
  • Likewise for bottles A, D, F, and G.
Let us now concentrate on B, C, and E, which must have prices $3, $4, and $5 -- not respectively, of course.  Going back to the original equations, we recall that...
A + F = C + E and, since A + F = 8,
C + E = 8, but...
...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...

    B = $4.
Give that a ho-hum. We now have figured out the price of merely one of the bottles. All that effort to go from seven unknowns to six.  Fortunately, the six unknowns are reciprocious in pairs.  The sophisticated solver will see at once that there are only eight combinations as follows:
 
A
F
C
E
D
G
2
6
3
5
1
7
2
6
3
5
7
1
2
6
5
3
1
7
2
6
5
3
7
1
6
2
3
5
1
7
6
2
3
5
7
1
6
2
5
3
1
7
6
2
5
3
7
1

That'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!
A + B = D + E is still satisfied if A = 6, F = 2, C = 5, E = 3, D = 7, and G = 1...
...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!
A + C = B + D is still satisfied if A = 6, F = 2, C = 5, E = 3, D = 7, and G = 1...
...and so it goes. Seems there are two solutions to this puzzle.
 
A
B
C
D
E
F
G
$2 
$4 
$3 
$1 
$5 
$6 
$7



OR



$6
$4
$5
$7
$3
$2
$1

Don't you wonder what the prize is for guessing the prices for those silly bottles?
 


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