13
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Shown below is 6x5 grid drawn on a paper with various numbers (1 to 9) and letters filling the 30 boxes.

Top row: V, 4, 6, L, 1, 5. Second row: 2, 8, 3, 1, 8, 9. Third row: 6, 4, C, 8, 5, 3. Fourth row: 4, 1, 9, 3, 2, 7. Fifth row: 3, X, 8, 1, 5, D.

By folding the paper grid 5 times or less can you get to a 3x3 grid showing only the numbers 1 to 9. The folded paper should look like

3-by-3 grid of empty boxes

with the numbers 1 to 9 that you can read directly. All numbers must show up on the 3x3 grid with 1 number per box. Order not necessary. The answer with the least number of foldings will be accepted. Folding only.

Some creative thinking may be needed.

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1
  • $\begingroup$ Yes. Back is blank. I is one. $\endgroup$
    – DrD
    Dec 15, 2021 at 15:58

1 Answer 1

6
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I believe I have a solution that uses

4

folds.

The key to this puzzle is to realize that:

many of the digits can be inverted and still be valid. This is true for 1, 2, 5, and 8, while 6 and 9 will be valid but swap values. Additionally, while most of the letters cannot appear in the final grid, the L is special, as it looks just like a seven when flipped. Thus, we can use it in place of the seven, which would otherwise have forced us to use the only 7 in the grid. This allows us to create solutions where some of the cells are "upside-down", but will still appear as a valid digit.

Folds:

folds
Where all folds are away from the viewer.

Result:

If we look at the resulting paper, we end up with:
wrong
Which clearly isn't the solution.

However, if we flip the paper around, we end up with:
right
Which looks a lot more like our desired result.

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3
  • $\begingroup$ Jung n terng jnl bs nccebnpuvat naq fbyivat. Irel jryy qbar! $\endgroup$
    – DrD
    Dec 15, 2021 at 16:00
  • $\begingroup$ Lbh fubhyq nyfb zragvba gung 9 vf vairegrq 6! $\endgroup$
    – DrD
    Dec 15, 2021 at 16:01
  • $\begingroup$ @DrD I added a bit more explanation about the nature of rot13(vairegrq qvtvgf). $\endgroup$
    – Surma
    Dec 15, 2021 at 16:07

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