58th Ukrainian National Mathematical Olympiad

Consider all possible tens of neighboring military ships. There exist ten neighboring military ships such that there are at least 4 destroyers among them. The first battle ship can hit these 10 targets. Consider the rest 20 targets and divide them into two groups: with odd position and with even position. One of these groups includes at least half of the destroyers that are saved after first rocket launch. So the second battle ship can hit this group. Therefore, if the first battle ship hits $k\ge 4$ destroyers, then the second hits at least $\frac{1}{2}(10-k)$. And in total battle ships hit $\frac{1}{2}(10-k)+k=5+\frac{1}{2}k\ge 7$ destroyers.

Now we build an example how to place military ships in a way that at least 3 destroyers will be saved. For this purpose we place ships such that every third is a destroyer. Then the first battle ship will hit at most 4 destroyers. If it is exactly 4, then the second battle ship could not hit more than 3, because there is the same amount of destroyers on odd and on even positions. If the first battle ship hits 3 destroyers, then the second could not hit more than 4.