100 Soldiers riddle

Solution 1:

Here is a way of rewriting your original argument that should convince your friend:

Let $A,B,C,D\subset\{1,2,\dots,100\}$ be the four sets, with $|A|=85$,$|B|=80$,$|C|=75$,$|D|=70$. Then we want the minimum size of $A\cap B\cap C\cap D$. Combining the fact that $$|A\cap B\cap C\cap D|=100-|A^c\cup B^c\cup C^c\cup D^c|$$ where $A^c$ refers to $A$ complement, along with the fact that for any sets $|X\cup Y|\leq |Y|+|X|$ we see that $$|A\cap B\cap C\cap D|\geq 100-|A^c|-|B^c|-|C^c|-|D^c|=10.$$

You can then show this is optimal by taking any choice of $A^c$, $B^c$, $C^c$ and $D^c$ such that any two are disjoint. (This is possible since the sum of their sizes is $90$ which is strictly less then $100$.)

Solution 2:

If you add up all the injuries, there is a total of 310 sustained. That means 100 soldiers lost 3 limbs, with 10 remaining injuries. Therefore, 10 soldiers must have sustained an additional injury, thus losing all 4 limbs.

The manner in which you've argued your answer seems to me, logical, and correct.