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Here is a challenge problem my math teacher gave to his pre-calculus class. I saw it. I attempted it. And I failed. It's sort of bothering me because my teacher said, "It's simpler than you're making it out to be." That sort of bothered me even though I know he's right. Anyways, time for the problem. It is as follows:

"A pumpkin growing contest occurs every year in Half Moon Bay. The pumpkins are weighed to see who grew the heaviest one.

One individual decided to weigh his 5 pumpkins two at a time. He recorded the following waits: 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. How much did each pumpkin weigh?"

This is my attempt.

Let $a_1, a_2, a_3, a_4, a_5$ be the weights of the pumpkins in increasing order, i.e. such that $a_1 \le a_2 \le a_3 \le a_4 \le a_5.$ If we take the sum of the 10 distinct weighings then we will have four times the sum of the weights of the pumpkins. In other words, $$110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1156$$ which is equivalent to, $$(a_1+a_2)+(a_1+a_3)+(a_1+a_4)+(a_1+a_5)+(a_2+a_3)+(a_2+a_4)+(a_2+a_5)+(a_3+a_4)+(a_3+a_5)+(a_4+a_5).$$ So we have $$ 4(a_1+a_2+a_3+a_4+a_5) = 1156$$ $$a_1+a_2+a_3+a_4+a_5 = 289.$$

Since $a_1, a_2$ weigh the least their sum must weigh the least, so $a_1+a_2 = 110.$ By similar reasoning we can see that $a_4+a_5 =121.$ Now we can see that $a_3=58.$ This ends my attempt.

I tried to go further but I couldn't. One person told me that it could be solved by system of equations but I thought that systems of equations might be troublesome because the problem does not specify which pairs of pumpkins correspond to which weights. But I don't doubt that it can be solved with systems of equations.

Any solutions would be much appreciated! Many thanks to all!

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You have already spotted that $a_1 + a_2 = 110$ and $a_4 + a_5 = 121$. But we also have $a_1 + a_3 = 112$ and $a_3 + a_5 = 120$.

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    $\begingroup$ Well, this is embarassing... I thought about that but somehow I convinced myself that that wasn't right. Such is life. Thank you! $\endgroup$ – Saudman97 Sep 13 '14 at 2:14
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Note that $a_3=a_2+2$ and $a_3=a_4-1$.

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