Skyscrapers 1: Elementary

Let a_1 < a_2 < \ldots < a_5 be the number set. The clue above the a_5 in the top row must equal a_5, and the clue above the a_4 must equal a_4 + a_5. If you imagine a fifth clue written in for the leftmost column, then the same goes for this column’s a_4 and a_5 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 a_5 and a_4 + a_5 start with the same digit (E), we have a_4 \le 9. Since there are four distinct initial digits among all clues, we must have a_4 = 9, a_3 = 8, a_2 = 7, a_1 = 6, and a_5 = 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: