This is in continuation of
Pumpkins and
Pumpkins 2.
Six pumpkins - each having a different weight - are weighed two at a time in all 15 sets of two. The weights are recorded as: 27, 36, 39, 45, 48, 51, 54, 57, 63, 66, 72, 75 and 81 pounds.
- Precisely two of the values occurred exactly twice, but these values were only written down once.
- Each of the individual weights (in pounds) is a positive integer.
Derive the weights of the pumpkins and which two values occur twice.
Well first, note that all the partial sums are divisible by 3. This means that all of the pumpkin weights are divisible by 3.
To simplify the math, let's divide everything by 3. (We'll multiply by three after solving this modified problem).
Then, the new totals are:
9, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 27
Let the pumpkin weights (smallest to largest) be a, b, c, d, e, f
Then 9 = a + b
12 = a + c
13 = a + d, or b + c, or both
If 13 = b + c, then {a,b,c} = {4,5,8}
If 13 = a + d, then {a,b,c,d} = {4,5,8,9} or {3,6,9,10} or {2,7,10,11} or {1,8,11,12}
Similarly,
27 = e + f
25 = d + f
24 = c + f, or d + e, or both
If 24 = d + e, then {d,e,f} = {11,13,14}
If 24 = c + f, then (c,d,e,f} = {9,10,12,15} or {8,9,11,16}. We need not consider further, since c >= 8 (from analysis above)
Putting these two results together, {a,b,c,d,e,f} must be either
(1) {2,7,10,11,13,14} or
(2) {3,6,9,10,12,15} or
(3) {4,5,8,9,11,16} or
(4) {4,5,8,11,13,14}
We can rule out (1) because 10 + 13 = 23, which is not a possible sum
We can rule out (3) because 9 + 11 = 20, which is not a possible sum
We can rule out (2) because there is no way to achieve a sum of 17
Therefore, the only possible solution is 3*{4,5,8,11,13,14} = {12,15,24,33,39,42}
Charlie has already verified that this solution works.
Edited on January 23, 2014, 12:22 am
Analytical Solution (spoiler)
