7 fishermen caught exactly 100 fish and no two had caught the same number of fish. Then there are three who have together captured at least 50 fish. $7$ fishermen caught exactly $100$ fish and no two had caught the same number of fish. Prove that there are three fishermen who have captured together at least $50$ fish.


Try: Suppose $k$th fisher caught $r_k$ fishes and that we have
$$r_1&lt;r_2&lt;r_3&lt;r_4&lt;r_5&lt;r_6&lt;r_7$$
and let $r(ijk) := r_i+r_j+r_k$.
Now suppose $r(ijk)&lt;49$ for all triples $\{i,j,k\}$.
Then we have $$r(123)&lt;r(124)&lt;r(125)&lt;r(345)&lt;r(367)&lt;r(467)&lt;r(567)\leq 49$$
so  $$300\leq 3(r_1+\cdots+r_7)\leq 49+48+47+46+45+44+43= 322$$
and no contradiction. Any idea how to resolve this?
Edit: Actually we have from $r(5,6,7)\leq 49$ that $r(4,6,7)\leq 48$ and $r(3,6,7)\leq 47$ and then $r(3,4,5)\leq r(3,6,7) - 4 \leq 43$ and $r(1,2,5)\leq r(3,4,5)-4\leq 39$ and $r(1,2,4)\leq 38$ and  $r(1,2,3)\leq 37$ so we have:
$$300\leq 49+48+47+43+39+38+37= 301$$
but again no contradiction.
Let's work with the lowest four numbers instead of the other suggestions.

Supposing there is a counterexample, then the lowest four must add to at least $51$ (else the highest three add to $50$ or more).

Since $14+13+12+11=50$ the lowest four numbers would have to include one number at least equal to $15$ to get a total as big as $51$.

Then the greatest three numbers must be at least $16+17+18=51$, which is a contradiction to the assumption that there exists a counterexample.

The examples $18+17+15+14+13+12+11=100$ and $19+16+15+14+13+12+11=100$ show that the bound is tight.
a = 0
b = 1
c = 2
d = 3
e = 4
f = 6
g = 84
In other words no mathematical solution to this riddle since there are many possible answers