This is part 14 of the puzzle series Around the World in Many Days. Each part is solvable on its own.

Deаr Puzzling,

The larger grid is a Fillomino puzzle. Each of the six different letters in the grid (F, K, O, R, S and T) stands for a different number 1–6. Divide the grid into orthogonally connected regions so that the number in each cell indicates the size of its region. Multiple given numbers can be part of the same region, and there can be regions with no given number at all. Regions of the same size cannot share an edge, but they can touch at a corner.

The smaller grid is a Star Battle puzzle. Place one star in each of the five regions in the grid so that there are no two stars on the same row, in the same column, or next to each other, even diagonally.

Today I have travelled along the coastline and arrived at an ancient fortified town at the end of an impressive mountain-rimmed bay. Can you guess where I am?

Love, Gladys.

Empty Fillomino grid
Empty Star Battle grid
Fillomino on Penpa+
Star Battle on Penpa+

Gladys will return in Paranormal Activity.

  • 4
    $\begingroup$ In case anyone's interested, this entire series is also being covered on my Youtube channel. $\endgroup$
    – Jafe
    Commented Mar 3, 2023 at 10:00

1 Answer 1


Gladys is in

Kotor, Montenegro.

First, let's solve the fillomino.

S or T must be 1 (because they block one another in the top-right corner). If S isn't 1, it must be at least 4 (because of the configuration in the top-left, remembering that K can't be 1), which would cause R to be 1 on the bottom-right. So S is 1.

K on the top-left is then 2.

R can't be 3, because of the configuration on the top-center. And if it's 4, that leaves a 1 next to the existing Ses there. So R there must be 5 or 6. But trying to make it 6 cuts off a 2-area T. So R is 5.

Because O on the top-right can't be 1 or 2, T there is limited to 3. O is then ≥4 which means the R in the right-center must expand downwards enough to meet the R on the bottom.

Trying to make O be 4 and F be 6 doesn't allow the two Os on the bottom-left to coexist. So O is 6 and F is 4, and the rest of the board fills easily.

For the star battle,

the L shape on the top-left must have its star in the second column (since the first column is taken by the bottom-left shape). So the second column and first row are both taken. That means the two shapes on the top-right must have their stars in columns 3–5, rows 2–3; then neither has it in column 4 to avoid adjacency. The rest follows easily.

There are a "T" and a "K" just outside the star-battle grid, implying we should find a five-by-five subgrid of the fillomino which has "T" and "K" in those positions outside it. There's only one.

On doing so, the stars spell out "Kotor", which is indeed "an ancient fortified town at the end of an impressive mountain-rimmed bay".


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