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Backstory contains important information, but nothing hidden. Also, tl;dr parts are bolded, but I can't guarantee I didn't forget bolding some part.


Recently there's been a craze here on Puzzling, so I thought I'd try my hand on them. After some serious considerations, I decided to buy myself a slitherlink book, because all you've to do is to draw lines and count up to 3; doesn't sound that hard, right? But of course, I wasn't going to spend all my money on some puzzle book, so I bought the cheapest one they had at the store.

And boy, it's the worst puzzle book ever. The characters are barely readable, the pages are coming off, and the puzzles are downright ridiculous.

Why, take the very first puzzle (I started with easy ones). It was a 5x5 slitherlink, and the weird thing is every cell had a clue-number in it! Seriously, that's basically reducing the puzzle to a child's play; what moron would need all 25 clues? On the top of it, it didn't even have a unique solution! And there was some nonsense fluff at the bottom; probably a lame attempt at making up for the lack of actual puzzle content.

The worst shock was yet to come, though. When I finished the slitherlink (barely took a minute), I accidentally spilt water on it, and all but three numbers in the grid got just washed off! Guess they used the cheapest ink they could get for printing. Too bad, because I hadn't even started trying to make sense of the stuff at the bottom.

puzzle

Now I don't remember what the actual puzzle was; what I do remember, though, is that the puzzle had at least two distinct solutions. For one thing, at least one of the three remaining clue-numbers could be satisfied in two different ways.

Can you help me reconstruct the slitherlink puzzle and find out the final, five-letter answer?

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  • 4
    $\begingroup$ I saw the title and prepared to flag this as spam :-P $\endgroup$ – Rand al'Thor Jul 14 '17 at 15:10
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The final grind and its two solutions:

solution 1 solution 2

From here

we can calculate the sums of the columns, treating each row as a five digit number. The sum is 129910.

By splitting the number as indicated by the dotted lines, we get the numbers 1, 2, 9, 9, 10.
The corresponding letters (A -> 1) are A, B, I, I, J

Now

adding the letters S, H, Y, E, R, but counting from A = 0 (S = 18, H = 7, ...) we get the final solution: SIGMA

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