19
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... seriously, this is definitely NOT that "kind of puzzle"! ;)

So let me create this puzzle to freshen you up!

It is inspired by a meta question: Yet another 2018 puzzle.


2018

Put the digits 2, 0, 1, and 8; exactly once for each grid color -- excluding black -- (some of them are already put and denoted by a black font-color) so that there is at least one row/column which has the numbers whose sum is 0, 1, 2, 3, and so on until 11.

There are also additional numbers denoted by red font-color which will be included on the sum of its row and column. The row or column that doesn't have any number will be having null (not zero) sum.

Clarifications: There are 5 rows and 10 (err.. 11) columns here, where the first row currently has the sum of 7 (-1 plus 8) and first column has null.

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  • 2
    $\begingroup$ LOL at quickly heading off people's dread at seeing '2018' in the title of a puzzle! :-) $\endgroup$ – Phylyp Jan 24 '18 at 10:09
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    $\begingroup$ I came here to downvote, but you get a +1. $\endgroup$ – prog_SAHIL Jan 24 '18 at 16:26
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Solution:

enter image description here

Method:

Not really much to say here other than some observations to start:

- The number 1 (blue cells) will always have a total of 11.

- The -2 and -1 in the first and last row are very valuable to make 6 and 7, which would otherwise be very expensive to make. We'd be wise to use them for this, which means top row is off limits and bottom row is reserved for the 8 from the number 1

- Since all columns are made with the same colors, and there are no duplicates allowed within the same colored cells, the only way we can get 4 and 5 are using the rows. 4th row is ineligible for this, so the 2nd and 3rd rows are reserved for those two totals.

- The only remaining row (4th row) is the dump row. All other rows have a specific target totals already, so this row will be where we dump any excess numbers. Mainly those pesky 8s that are too large to be useful.

These observations are enough to derive the rest from brute force. Play with the position of 1s and 2s so that we get 0..3 and 8..10 from the columns, and voila!

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  • 1
    $\begingroup$ Really interesting observations. I'd got no clue where to start. Once you see them they feel obvious but I managed to miss every one of them (including the one on the 1/blue cells). :) $\endgroup$ – Chris Jan 24 '18 at 14:01
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    $\begingroup$ Great job! :D The main observation is actually to know that the possible sums for column are {0, 1, 2, 3, 8, 9, 10, 11}; leaving sums of 4, 5, 6, and 7 from the row part. Indeed ndeed, -1 and -2 is used for creating 6 and 7 (by using 8) and then the rests are not pretty trouble-free to fill once you got all of these steps! :) $\endgroup$ – athin Jan 24 '18 at 14:34

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