Carrying weight of $270$ kg Problem: A total of $270$ kg of watermelons, each weighing at most $7$ kg, must be transported by $11$ carriers at once. Show that if each carrier can carry up to $30$ kg at a time, this can be done regardless of the weight of the individual watermelons.
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My opinion about the solution: Let the number of watermelons be $n$. Let the weight of the watermelons be $0< x_1 \leq x_2 \leq \dots \leq x_n \leq 7$. Since $$\sum_{k=1}^{n} x_k = 270 $$
and $7n \geq n\cdot x_n \geq 270$ we find $n \geq 39$.
For $39 \leq n \leq 44$, $\{ x_1, x_2, x_3, x_4\},  \{ x_5, x_6, x_7, x_8\}, \dots, \{ x_{41}, x_{42}, x_{43}, x_{44}\}$ groups can be transported by $11$ carriers.
For $n>44$, I have no enough idea. For sufficiently large $n$ values, some weights of watermelons will decrease. Thanks for your interest...
 A: Assume that we distribute the melons as follows. Sort them by decreasing weight (starting from the heaviest), and always give the next melon to some carrier who has sufficient unused carrying capacity. I claim that this algorithm always succeeds in distributing the entire load to the eleven carriers.
Assume contrariwise that we fail to assign the melon number $n$, weighing $x$ kilos to some carrier. For this to happen we must have already assigned each carrier a load of more than $30-x$ kilos. If $w$ stands for the combined weight of the already assigned melons, we arrive at the inequality $w>11(30-x)=330-11x$.
On the other hand, because at least $x$ kilos remain, we have assigned at most $270-x$ kilos, and hence $w\le 270-x$ (with equality only if this is the last melon).
Putting these pieces together we get
$$
330-11x< w\le 270-x,
$$
from which we (ignoring the temporary dummy variable $w$) immediately solve $x>6$. Recalling that the preceding $n-1$ melons all weigh $\ge x$ kilos, we have at least $n$ melons weighing over $6$ kilos each. Because $nx\le 270$, it follows that $n\le 44$. But the OP's argument shows that the distribution of at most $44$ melons, each weighing at most $7$ kilos, poses no problems in general, and for our algorithm in particular. Therefore the problematic situation cannot arise.

It is worth pointing out that $270$ kilos is the maximum total load the group of carriers is guaranteed to be able to haul. Inspired by the argument we see that if a total load of $270+\epsilon$ kilos consists of $45$ melons, each weighing $6+\epsilon/45$ kilos, then the task becomes impossible. For $45>4\cdot11$, so by the pigeonhole principle at least one carrier needs to haul at least $5$ melons, causing their load to become $30+\epsilon/9$ kilos, which is a bit too much.
