If you liked my previous puzzle, here's more.
Two puzzles. Three giveaways. Solve puzzles to find the 3rd giveaway.

Also check out my other digits puzzle.

Puzzle has ended. Solution is posted here.

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7 years ago*

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seriously? again? xD
*sigh, looks like i have to work on my code again xD

7 years ago
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You got me this time. Have to go out now.

I'll solve it later

Out of time to solve the 3rd puzzle.

7 years ago*
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Bump for all solved

7 years ago
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Bump for everything solved :) *happy

7 years ago
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Well done!

7 years ago
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Bump ! it was pretty hard and also nice puzzle for me. thanks :)

7 years ago
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BUMP for solved

7 years ago
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people already solved it, but I got a solution where I got one number in two different threes (so 6 remaining)

7 years ago
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You need to find a solution where every number is in one three only.

7 years ago
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found it
btw there are at least 5 threes with that sum, so had to look for the 3 with unique numbers

7 years ago
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Bump for solved!

7 years ago
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Solution:

  1. 4+139+146 = 83+102+104 = 90+98+101 = 289. Leftover characters: xXJhs
  2. 5+107+193 = 49+73+183 = 70+112+123 = 305. Leftover characters: aUiYa

Both giveaways stated that for the third giveaway you need numbers 101 4 193 107 70. The trick was to find characters corresponding to these numbers on the puzzle diagrams. You didn't even had to solve them to find that:

  • From the first puzzle: 101 => O, 4 => m, 70 => s.
  • From the second puzzle: 193 => 9, 107 => V, 70 => s.

Knowing this, numbers 101 4 193 107 70 could be translated to Om0Vs.

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7 years ago
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