Let be the number set. The clue above the in the top row must equal , and the clue above the must equal . If you imagine a fifth clue written in for the leftmost column, then the same goes for this column’s and as well.

So, the top row and the leftmost column must have two clues in common. Could one of them be RY? No, because then the rightmost-column clue would also equal RY.

Therefore, one of the common clues is EM and the other one is the written-in imaginary clue. That clue, though, would then equal the clue above the top left cell, or EL.

Since and start with the same digit (E), we have . Since there are four distinct initial digits among all clues, we must have = 9, = 8, = 7, = 6, and = 10E, and A, R, and T must equal E + 1, E + 2, and E + 3 in some order.

Additionally, in all lines with an E + 3 initial digit, five buildings must be visible, in all lines with an E + 2 – four, and in all lines with an E + 1 – three. It is not difficult to complete the solution from here: