Solve this hexagon puzzle

Question

I recently stumbled upon this interesting puzzle.

I wanted to know how to tackle such problems. Basically, the intuition or the thought process while solving such problems.

Source of the puzzle

Solution

Thanks.

Answer 1

1. Blue 3059 implies the green 3059. Of course $3059=23\cdot 19 \cdot 7$, so there must be black 23, 19 and 7.
   The only place having at least 23 tiles in sight is in the centre, so we must place black 23 there. Also in sight of black 23 and blue 3059 must be green 3059.
   All tiles in sight of black 23 becomes red except one that can be potentially green 3059.
2. Among the fields that remain empty there are only 5 with at least 19 tiles in sight. Only one have in sight only red or empty fields and one potencially containing green 3059. Place the black 19 there and green 3059 in adequate place. All remaining tiles in sight of black 19 becomes red.
3. Green 7 imples the black 7. There is only one empty tile having in sight at most 7 red tiles - place the black 7 there. The remaining fields in sight of black 7 can not be red. In sight of the green 7 can not be more black fields.
4. Blue 121121 imples green 121121. $121121=7\cdot 11^3\cdot 13$. The green 121121 bust be in sight of the only black 7 and blue 121121 - there are only 3 such fields. Two of them doesn't have enough clear tiles to place the remaining black numbers, so we can place green 121121 in sufficient place.
   Our green 121121 have in sight 3 tiles that can be black - we must place in them two 11s and one 13. It is easy to see, that one of them already have 12 red tiles in sight, so we place black 13 there and remaining 11s in two remaining tiles. Colour two tiles next to 13 and 11 red.
   Green $847=7\cdot 11^2$ have in its sight black 7 and two black 11s - in its sight there should be no more black tiles.
5. Red 0s are blocking blue numbers in sight.
   Red 843 have in its sight only one tile that can be blue - put there blue 847.
   Blue 847 requires green 847 - there is only one tile in sight of blue 847 that can be green - put green 847 there.
   Blue 7 requires green 7 - there are two tiles that can be green, but only one gives 7 - put green 7 there.
   Red 3997 requires 3150 in blues - put blue 3059 and blue 91 in sufficient tiles.
   Blue 91 requires green 91 - there is only one tile where we can put green 91.
   There is one more place, where could be green 77, but it will collide with blue 7. Just leave this place empty.
6. Now it's just filling red tiles with numbers