My friend William and I wanted to make a word search (number search?) of five digit numbers in a 5x5 grid, filling all spaces to form 12 clues.

To simplify things, we decided we would add a single leading zero if the value was less than one, drop any decimal symbol, and truncate at 5 digits if needed. So, for example, Catalan’s constant should be depicted as ‘09159’ instead of 0.9159.

He grabbed a piece of paper and set to work, filling in two five digit numbers in the grid. And then (to my dismay), he announced “this is hopeless” and left me to finish alone.

I didn’t pay attention to his numbers except where mine intersected his, and happily filled in the remainder of the grid.

Only when we’d finished did I spot something problematic.

The question is this: which two clues did William fill in and how do you know?

Below are the 12 clues. I'm sorry they are no longer in order. Like I said, they kinda got a little messed up. I’m sure you’ll sort it out.

To ensure this isn’t TOO frustrating, here is a starting digit.


THE WORD-SEARCH CLUES (In no particular order)

  1. Nickel-niobium-beryllium
  2. SQRT(2)
  3. π
  4. e
  5. Klaus Teuber game played around LA
  6. SQRT(559)
  7. Alpha (Feigenbaum's specifically; no need to make this puzzle harder than it already is)
  8. SQRT(449)
  9. 886.., er, I can’t remember the last 2 digits
  10. x (e.g. a random 5 digit integer)
  11. x/2691
  12. SQRT(968)

Good luck!

Note: As mentioned in commentary, both x's are the same value. It doesn't really matter too much what 'x' is, since 1.2345=12.345=12345 etc., and on top of that, the other 10 clues will give you enough information to solve this, but if it helps as a checksum, whoever filled it in used a 5 digit integer as 'x'. So I've updated the clue.

  • $\begingroup$ x is the same in both 10 and 11, correct? $\endgroup$ – Rob Watts Mar 25 at 21:44
  • $\begingroup$ Yep I'm not that cruel ; ) $\endgroup$ – Amoz Mar 25 at 21:46

The answer is that William had clues

5 and 10, and they are written upside-down relative to the other clues.

To start with, here's what we can initially determine about the numbers for each clue:

  1. Nickel-niobium-beryllium

I believe we should use the atomic numbers for these elements, giving us 28 41 4

  1. SQRT(2)

The square root of 2 starts with 1.4142 -> 14142

  1. π

3.1415926... -> 31415

  1. e

2.718281... -> 27182

  1. Klaus Teuber game played around LA

Klaus Teuber is the creator of Settlers of Catan, and putting Catan around LA gives CataLAn - Catalan's constant of 09159 mentioned in the question. Thanks @Braegh for pointing out the hint I missed.

  1. SQRT(559)

23.64318... -> 23643

  1. Alpha (Feigenbaum's specifically; no need to make this puzzle harder than it already is)

https://en.wikipedia.org/wiki/Feigenbaum_constants#The_second_constant - 2.5029078... -> 25029

  1. SQRT(449)

21.189620... -> 21189

  1. 886.., er, I can’t remember the last 2 digits

  2. x (e.g. a random variable)

  3. x/2691

  4. SQRT(968)

31.112698... -> 31112

Some thoughts about how we can figure out where they go in the grid:

Four clues intersect in the middle, so that digit must be the middle digit of four clues. Five clues have 1 as their middle digit and two clues each have 4 and 6 as their middle digit. However it is possible for clues 10 and 11 to both have 4, or 6 as their middle digits, which makes things a little more complicated.

Taking the cases one by one:

The center digit can't be 6. It requires the clues for 10 and 11 to go through the center, along with 23643 and 886??. However, this leaves the first 3 of 23643 without anything that can possibly intersect with it.

The center digit can't be 4. For 31415 to not intersect the 3 in the lower left, one of the other clues starting with 3 would have to have a 3 or 5 in its 3rd or 5th digit. With 31415 in the diagonal with the 3 at the lower left, 28414 can't fit - vertically or horizontally requires 31112 or 23643 to have a 2 or 4 as the third digit, while diagonally would only work with one of these changes: 28414->28412, 31112->31114, 23643->43643.

So we now know one more digit in the grid:

The center digit must be 1.

Now we can determine one of the numbers that must pass through the 3 in the lower left:

23643 must intersect with it. There are 6 cases to consider where it doesn't, but none of them are very hard to disprove. I'll avoid having my answer get too bloated and leave that as an exercise to the reader.

Here's what the grid looks like now:

6 1

We can make a significant observation at this point:

We do not have any numbers that fit the middle row. That means either clue 10 or 11 is 6?1?? (possibly reversed). Also, we have two clues left that start with 3, but two rows and a diagonal that need to have a 3 at the end. One of the biggest obstacles to placing numbers was not knowing if one of the wildcards could be in the way, but we now can get around that.

So the next two numbers to place in the grid are:

28414 is placed backwards in the fourth row - it is the only available clue that has a 4 at the end. Then the middle column must be 14142. Nothing would be able to fit in the top row if the number was written bottom-to-top, so it must be in there with the 1 at the top.

We can now make an observation similar to our previous one:

None of the remaining clues fit the 3?2?? in the bottom row, so that must be the other wildcard.

The grid at this point:

2   1
3 4
6 1
4 1 4 8 2
3 2

It feels rather easy to place the clues now, leading to the following two steps (I added two numbers each time).

2   1   2
3 1 4 1 5
6 1
4 1 4 8 2
3 2

2   1   2
3 1 4 1 5
6 1 0
4 1 4 8 2
3 2 9

Here's where a bit of lateral thinking comes into play

09159 doesn't fit in anywhere that's left. Unless, that is, the author wasn't paying enough attention to realize that it was upside down in the middle row as 65160.

2   1   2
3 1 4 1 5
6 5 1 6 0
4 1 4 8 2
3 2 9

For the next two clues

886?? must be in the fourth column from bottom to top, and 27182 must be in the top row but we don't know if its backwards or forwards yet.

We're down to our last two clues, but not out of the woods quite yet

Clues 10 and 11 correspond to ?151? and 3?289. With a little searching of the number space (3?289 or 982?3, multiplied and divided by 2691 with a little fudging), I found that if x=81510, then x/2691=30.289, giving us our two numbers.

And here's the final grid:

2 8 1 7 2
3 1 4 1 5
6 5 1 6 0
4 1 4 8 2
3 0 2 8 9

We can also easily tell which clues William filled in and why:

William had clues 5 and 10 and are written upside-down relative to all the other clues. We already saw where clue 5 was written upside-down, and clue 10 is the only clue that can been written upside-down and still be the same. It's also not too hard to see that if clue 5 is not upside-down then generating the number search is indeed as hopeless as William believed.

  • $\begingroup$ 5 could be rot13(PNGNA nebhaq YN = PNGNYNA), perhaps? $\endgroup$ – Braegh Mar 25 at 22:46
  • $\begingroup$ X is a 5 digit integer if it helps, although that's not critical I believe $\endgroup$ – Amoz Mar 25 at 23:02
  • 1
    $\begingroup$ That's not how I remember $e$. $\endgroup$ – Bass Mar 25 at 23:45
  • $\begingroup$ @Bass ack, typo! Fortunately, I haven't needed to rely on it at all so far $\endgroup$ – Rob Watts Mar 26 at 0:09
  • 1
    $\begingroup$ @RobWatts the constant rot13(juvpu vf rkcyvpvgyl zragvbarq va gur chmmyr cbfg vgfrys!) $\endgroup$ – Braegh Mar 26 at 1:24

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