# 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: 