This puzzle proved fairly simple for solvers with experience solving skyscrapers puzzles. The key in solving both halves of this puzzle is starting with placing as many sixes as possible, taking advantage of the clues of value one or five. The large clues make restrict the possible values of certain spaces quite a bit. In the interest of brevity, I have hidden the step-by-step process here, but the completed grids are shown below:
Start by placing a 6 immediately adjacent to any 1-clues. This gives three 6s across the two grids.
We will now focus on the right grid, since it is somewhat easier. Since the 2-clue in the bottom row sees its 6 on the far right, there must be a 5 in the lower left to hide the 1-4.
The 5-clues on the top and left require a 6 on one of the two furthest spaces. The 6 in the lower right corner blocks off the last space in both cases, so we can place both 6s.
The 6s in the second and fourth rows are now the only ones missing. Since the one in the second row must be at least three spaces away, and that is the most distant available space, we place one 6 there and the other 6 in the second box of the fourth row.
In the first column, we are missing the numbers 1-4. The 4 cannot be placed in row three (which must contain a 1 or 2 to satisfy the 5-clue) or row five (which must contain a 1-3 to satisfy the 4-clue). It also cannot be placed in the second row, since that will force a 5 to be placed in the second column (to satisfy the 3-clue). However, this 5 will never satisfy column two (since the 2-clue will always see whatever the number in row one is, plus the 5 and 6). Thus the first column 4 must go in row four.
Because of the 5-clue in column four, the fourth row in that column must contain a 4 or 5. Because we just placed a row four 4, the missing number must be a 5.
Similarly, the third row, fourth column square must be a 4 or a 5. With the 5 we just placed, it must be a 4.
To satisfy the five-clue in row three, the first three boxes must be 1, 2, and 3, leaving the rightmost box as a 5.
In the second column, we have already discussed that the 5 cannot be placed in the second row, and the sixth already has a 5. If we place it in row five, it messes up the 4-clue, so it must be in row one.
Since every square to the left of the 6s in rows two and five must be visible, the column two 1 goes in the bottom row.
This means that the second and fifth rows start 2, 3 and 3, 4 in some order. This means that the 3 for row two definitely occurs within the first three squares, so it cannot appear in column four. Therefore, the top two squares in the fourth column must be a 1 and a 2, with the 3 at the bottom.
With that 2 placed, row two must begin with 3, 4 leaving 2, 3 for the fifth row.
The 5 in the sixth column allows us to end row two with 5, 1.
The other 1s and 5s in the grid allow us to finish the fifth row with 5, 1, and 4.
The sixth row is missing a 2 and a 4. If the 4 was in the fifth column, the 3-clue would be invalid, so it must be in the third column, leaving the 2 in the fifth.
The remaining six numbers can all be filled in using process of elimination. 2, 4, and 3 on the top, and 1, 3, and 2 in the fourth row.
Now we move back to the left grid. The 4-clue and 3-clue in the second row force the 6 to be in the fourth column.
The 5-clue on the bottom forces the square below the 6 to be a 4 or 5. However, the 5-clue on the right requires it to be 4 or less. Therefore, it must be a 4, and the squares below it must be 3, 2, and 1, with the 5 on top.
In the rightmost column, the 5 cannot be placed in row two (because of the 3-clue), row three (because of the 5-clue), or rows five or six (thanks to the bottom 4-clue). Therefore, it must be in row four.
Likewise, in the same column, the 4 cannot be placed in the third row (due to the existing 4), fifth row (because of the 4-clue on the side), or sixth row (because of the 4-clue on the bottom). Thus, it must go in the second row, which forces a 5 placement next to it.
Thanks to the 4-clue on the left, the second row must start with 1, 2, and 3.
Since there is already a 1 on the bottom row, the bottom-right corner must be a 2 (to satisfy the 4-clue). Likewise, the square above it must be a 3, and the remaining sixth-column square must be a 1.
The 6s in the third row (because of the 5-clue) and the fifth row (because of the 4-clue, plus the hidden 2) must be in the first two columns. However, thanks to the 1 we just placed in the first column, the first column 6 cannot be in row three. Thus, there are 6s in row three, column two and row five column one.
Thanks to the 5-clue on the right, there must be a 5 between the 4 and 6 in the third row.
The only way to satisfy the 2-clue in the bottom row is to place a 6 between the 2 and 1. This leaves only one remaing location for the final 6 in the fourth row, third column.
Existing numbers force the missing numbers in the third column to be a 2, 1, and 4, reading down.
The placements of the 4 and 5 in the fifth row are forced by the 4-clue on the right.
The bottom row’s missing numbers, 5 and 3, are placed by process of elimination.
In the first column, the 4 cannot be in the first row (thanks to the 3-clue) or the third row (thanks to the existing 4). Therefore it must go in the fourth row, with the 3 on top and the 2 in row three to fulfill the 3-clue.
The remaining numbers (4 and 1 in the first row, 3 in the third, and 1 and 2 in the fourth) are all placed by process of elimination.
Solvers will notice that each of the numbers 1-6 are assigned to a meaningless phrase of twelve letters. The grids contain each number twelve times. Replacing each number, left-to-right, with the letters of its phrase gives the following grid:
The flavor text mentions going down and up, up and down. Reading the diagonals in that order gives LIFT FOR BRITS IS WHAT FOR USA. What the British refer to as a lift is called an ELEVATOR in the United States.
