The Drowning City

It follows from the rules of symmetry given in the first paragraph that $A_{1}=A_{4}=D_{1}=D_{4}$ and $B_{2}=B_{5}=E_{2}=E_{5}$ , giving this right off the bat:

Next we note that if 5 people are rescued on the right side of row $2$ and $E_{2}=3$ then $F_{2}$ must be either $2$ or $5$ . It can't be $2$ as then $3$ and $2$ would be next to each other and as they have a difference of $1$ this is not allowed, hence $F_{2}=5$ . Similar logic shows that for $12$ to be rescued from the top of column $E$ , $E_{1}$ must be less than $3$ , as it's next to $3$ and can't be $2$ , $E_{1}$ must be $1$ . Using the rules for similar rows it is given that $F_{2}=F_{6}=C_{2}=C_{6}$ and $E_{1}=E_{4}=B_{3}=B_{6}$ . Now we have

To rescue $12$ looking down $E$ it is now apparent that $E_{3}=8$ . By symmetry $E_{3}=E_{6}=B_{1}=B_{4}$ , giving us:

Looking at $F_{3}$ it can't be $1$ , $3$ , $5,$ or $8$ as it is around all of those. Since it is directly next to $8$ and $5$ it also cannot be $4$ , $6$ , $7$ , or $9$ and $F_{3}$ must be $2$ . $F_{3}=F_{6}=C_{1}=C_{4}$

By the same method, it is trivial to show $D_{3}$ to be $4$ and $A_{2}$ to be $9$ and because $D_{3}=D_{6}=A_{3}=A_{6}$ and $A_{2}=A_{5}=D_{2}=D_{5}$ our grid now looks like

Lastly use the same method of examining the squares around it to rule out possible values on $C_{3}$ to find it must be $7$ . $C_{3}=C_{6}=F_{1}=F_{4}$ so our completed grid is:

Question 1: $B_{6}<B_{5}<B{4}>B{3}$ so the only $B_{6},B_{5},B{4}$ could be rescued here, however note that because $A{4}<B{4}$ , $B_{4}$ was already rescued from the left of row $4$ so the answer is just $B_{5}+B_{6}=3+1=4$ , meaning $A_{1}=1$

Question 2: Going around the whole square and using the method of rescue that saves all buildings until a building is too tall to see the next rescues the following blacked out buildings

Summing the remaining buildings values gets an answer of $25$ meaning $A_{2}=3$

Question 3: Note that every diagonal has $3$ as the second building in it. This means a diagonal would only save additional people if it's corner is less than $3$ . The only corner that meets this need is the bottom right with a value of $2$ and this diagonal would save the following two boxes

The sum here is $9$ so $A_{3}=1$

The answer is then $\overline{A_{1}A_{2}A_{3}}=\boxed{131}$

Not a complete solution:

Start filling the 6x6 grid based on properties of the grid cells.

Start with given knowledge that $A_{1} = A_{4} = D_{1} = D_{4}$ and $B_{2} = B_{5}$ and $B_{2} = E_{5}$ and $B_{5} = E_{2}$ .

Then fill value of $E_{1}$ based on the knowledge of its neighbors and the fact that $12$ people will be saved when flash is above column $E$ .

Continue this process to fill out the grid. Afterwards compute the number of people saved by the team during its anticlockwise rescue along the outskirts of the city..

Note: Remember that people already rescued from a building can't be rescued again because they are already on the boat.

I think its way easier to solve this question, the tougher part is to understand the question in the first place.