# Every year, there is a contest…

Every year, there is a contest to see who has the heaviest pumpkins for that year.

Last year, a farmer brought 5 pumpkins to the contest. Instead of weighing them one at a time, he informed the judges,

"When I weighed two at a time, I got the following weights: 108, 112, 113, 114, 115, 116, 117, 118, 120, and 122."

How much did each pumpkin weigh?

I tried solving this problem by summing all the weights (1,155) and dividing by 10 to get the average weight of two pumpkins (115.5). I then divided by 2 to get the average weight of each pumpkin (57.75). To determine the middle pumpkin, I multiplied by 5 to find the total weight of the five pumpkins (288.75) and then subtracted the lightest and heaviest weight ('C' = 58.75). Fast forward, I determined 'A' = 53.25, 'B' = 54.75, 'C' = 58.75, 'D' = 60.75, and 'E' = 61.25.

However upon review, I determined that these values could not produce the 10 weights. In addition, I concluded the values must end in .5 to increase the combinations.

The question is: is there a correct answer and what is it, or is this unsolvable?

Thanks.

• In the 10 combinations, how many times is each pumpkin repeated ? List out the combinations and you will have your answer. – Shailesh Aug 26 '15 at 2:20
• The formula $2A=(A+B)-(B+C)+(C+D)-(D+E)+(E+A)$ might be useful, if you know which sums are which. (It follows because, distributing, we have $A+B-B-C+C+D-D-E+E+A$, and everything cancels out with its neighbor save for $A$.) – Akiva Weinberger Aug 26 '15 at 2:27
• Using my values — 'A' = 53.25, 'B' = 54.75, 'C' = 58.75, 'D' = 60.75, and 'E' = 61.25 — I was unable to provide combinations for 113, 115, 117, and 118. – Sentient Aug 26 '15 at 2:30
• @columbus8myhw I believe 'A' + 'B' = 108 and 'D' + 'E' = 122. These are the only sums I'm certain about. – Sentient Aug 26 '15 at 2:32
• I think it's unsolvable. – Akiva Weinberger Aug 26 '15 at 2:48

First, you’re right in thinking that the total weight is $288.75$; you can also get this by noticing that each pumpkin is weighed four times, and $\frac{1155}4=288.75$. However, the data are inconsistent.

The low total of $108$ must be the sum of the two lightest weights, so the three heaviest pumpkins altogether weigh $288.75-108=180.75$. The two heaviest together weigh $122$, so the middle pumpkin weighs $180.75-122=58.75$. Each pair of pumpkins has an integer total weight, so each of the other four weights must have a fractional part of $0.25$ – but that ensures that the sum of any two of their weights is not an integer.

• Moral of the story: If you're going to hold a pumpkin contest, be prepared to weigh the pumpkins yourself and don't trust the contestants. – JiK Aug 26 '15 at 15:09
• Along these lines, every one of the two-pumpkins weights is $288.75$ minus some three-pumpkins weight. So if the farmer had weighed the pumpkins three at a time, the resulting set of weights would be $\{180.75, 176.75, 175.75, 174.75, 173.75, 172.75, 171.75, 170.75,168.75, 166.75\}$. But each pumpkin’s weight is some weight-of-three-together minus some weight-of-two-together, and all such differences have fractional part $0.75$, which is impossible given the claimed two-pumpkins weights. – Steve Kass Aug 28 '15 at 1:08

Observations

Let the five terms be represented by $a<b<c<d<e$.

Let's suppose there are solutions to the problem.

On the lower end, we know that $a+b=108$.

We also can deduce that $a+c=112$

On the upper end, we know that $d+e=122$

We can also deduce that $c+e=120$

Substituting, we discover that $c=b+4$ and $c=d-2$.

Now we know the difference between our central 3 values. The center should be ~$b+3$

$$b=b$$ $$c=b+4$$ $$d=b+6$$

The range is $6$. As you already pointed out, the mean is $57.75$, which we'll say is $58$ to make calculations easier.

Using this, let's guess that $b$ is $3$ below $58$. When this is the case, our values are the following:

$$a=53$$

$$b=55$$

$$c=59$$

$$d=61$$

$$e=61$$

To achieve $b,c,$ and $d$, simply plug in $57$ for $b$ in the above calculations and you'll obtain $a$ and $e$ by subtracting $b$ from $108$ and $d$ from $122$, respectively. Obviously these cannot be the correct weights since $d=e$, but let's add each combination to see how close we are. Upon doing so, we get the following values:

$$a+b=108$$

$$a+c=112$$

$$a+d=114$$

$$b+c=114$$

$$a+e=114$$

$$b+d=116$$

$$b+e=116$$

$$c+d=120$$

$$c+e=120$$

$$d+e=122$$

Now we'll try guessing that $b$ is $3.5$ less than $58$

In this case, we get the following:

$$a=53.5$$ $$b=54.5$$ $$c=58.5$$ $$d=60.5$$ $$e=61.5$$

These values yield the following:

$$a+b=108$$ $$a+c=112$$ $$b+c=113$$ $$a+d=114$$ $$a+e=115$$ $$b+d=115$$ $$b+e=116$$ $$c+d=116$$ $$c+e=120$$ $$d+e=122$$

This set was extremely close, but we had duplicate $115$ and $116$ values and lacked our $117$ and $118$ values.

I know others have already shown that there is no solution, but I thought this might help show that although it cannot be done, there are pumpkin weights that yield almost all of your values.

Say we have $5$ different weighted pumpkins with weights $a, b, c, d, e$ such that $a\lt b\lt c\lt d\lt e$. There are $10$ different ways these pumpkins can be weighed two at a time. The different combinations are

1. $a+b$
2. $a+c$
3. $a+d$
4. $a+e$
5. $b+c$
6. $b+d$
7. $b+e$
8. $c+d$
9. $c+e$
10. $d+e$

Now a possible way to get the values of the weights is to make a system of equations. Since there are five unknowns we need five equations in order to be able to find a unique solution. Keep in mind that a system need not have a solution. Now our system has ten equations and in order for it to be consistent, i.e., have a solution, all 10 equations must be true. The problem is that we do not know exactly what all 10 equations must be. So we must find five equations that must be true and if that system has a solution, we will be able to recover all of the reported weights.

Now a few equations are obvious. Since $a,b$ are the smallest their sum must be the smallest. So we have $$a+b=108.$$ Similarly, since $d,e$ are the biggest, there sum must be the biggest, implying $$d+e=122.$$ Also, it can be seen that $$a+c=112$$ and $$c+e=120.$$ Already four equations! The fifth one can be obtained by summing all of the possible equations and then dividing by four on each side (which is what you and @columbus8mywh did in the comments) to obtain $$a+b+c+d+e=288.75.$$

This system has a unique solution. Namely, $a=53.25$, $b=54.75$ $c=58.75$, $d=60.75$, and $e=61.25$.

However, notice that this cannot be consistent with the entire system since if we were to weigh together, for example, $b$ and $d$, we would get a non-integer value. Since this is impossible the entire system is inconsistent and there is no solution.

As others have stated, you know the relative positioning of all the pumpkins from the problem.

Given an unknown a,

b=108-a c=b+4 d=c+2 e=122-d

This means if a is integer odd, they are all odd. If a is integer even, they are all even. Either case leads to every sum being even.

The only other alternative is that a is non-integer. In order for all possible sums to be integer, it must end in .5. If a is odd.5 then b,c,d are all even.5 and e is odd.5. If a is even.5, then b,c,d are all odd.5 and e is even.5.

In both cases, you get the following E-ven and O-dd sums (it's just an X-OR which is why it doesn't matter the evenness is flipped between the two cases):

EEEO OOE OE E

So it doesn't provide the requisite 3 odd sums and 7 even.

The farmer is a liar.