A number theory problem by A Former Brilliant Member

For the first bunker, let there be $x$ men. For the second, there must b $2x$ . For the third, there must be $4x$ . Summing these gets $7x$ men. Because there are a multiple of 5 dudes, let's try $x=5$ . $\frac{35-12}{35}<\frac{2}{3}$ , which clearly doesn't work. When we try $x=10$ , we see that $\frac{70-12}{70}>\frac{2}{3}$ . Therefore there were originally 70 men, and after 12 were eaten, there are $\boxed{58}$ men.

Let the total number of sailors initially be $S$ . Accordingly there are $x$ , $2x$ & $4x$ sailors in the first, second & third bunkers respectively. .: $S$ = $x$ + $2x$ + $4x$ = $7x$ . Given that $S$ is less than $100$ & multiple of both $5$ & $7$ , the only possibilities of $S$ are $35$ & $70$ . Now let us consider $S$ = $35$ . Out of which $12$ die. Remaining sailors are $23$ . But the number of survivors should be greater than $2/3$ of $35$ , which is approximately $23.33$ . Hence, $S$ = $70$ & not $35$ . The number of survivors is $70$ - $12$ = $\boxed{58}$ .