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Create a word search with the following conditions:

  1. Use the eleven words below. They can be spelled up, down, left, right, or any of the 4 diagonal directions.

ZERO
ONE
TWO
THREE
FOUR
FIVE
SIX
SEVEN
EIGHT
NINE
TEN

  1. The size of the word search must be 5 rows and 6 columns.

  2. The number of words that pass through each square must fit this pattern:

editing. one minute...

  1. The accepted answer will clearly and fully explain the flow of logic used from start to finish.
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  • 2
    $\begingroup$ Not necessarily "only" twice -- each particular '2' square is used for two words. Similarly the '3' squares. (might want to clarify that in the puzzle proper though) $\endgroup$
    – lorimer
    May 28, 2015 at 16:39
  • $\begingroup$ Do all words have to be written left to right? $\endgroup$
    – tfitzger
    May 28, 2015 at 17:05
  • $\begingroup$ @tfitzger See rule #1 $\endgroup$
    – JLee
    May 28, 2015 at 17:06
  • $\begingroup$ @JLee I'm curious, how did you come up with this puzzle? Even after knowing zero to ten could fit in a 6 by 5, it was still very challenging. I can't imagine figuring that out accidentally. $\endgroup$
    – Quark
    May 29, 2015 at 4:49
  • $\begingroup$ @Quark I was creating a different puzzle with these 11 words, then stumbled on this site: puzzlemaker.discoveryeducation.com/… and started playing around with the smallest possible word search that it could create with those 11 words. After over a hundred trials, the best it could do was 10 out of the 11, so I thought maybe it was impossible in a 6x5 grid. So, I saved a board without ZERO, and a board without ONE, ... up to TEN. Then I started to analyze the boards, to see if I could somehow manually add the missing number $\endgroup$
    – JLee
    May 29, 2015 at 12:36

1 Answer 1

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Here's my answer:

enter image description here

From this point on, there is no spoiler text, because it makes it easier to format, at least for me.

T E N I N E
F I O N E V
O G X R V I
U H H I E F
R T W O S Z

I started by figuring out the letter density in the overall puzzle. The letters break down as follows:

E   10
N   5
I   4
O   4
T   4
R   3
F   2
H   2
S   2
V   2
G   1
U   1
W   1
X   1
Z   1

Next I started working on the 3 squares. Since all three are in a diagonal, I started with the assumption that they are within the same word. This requires a word of length 5, which means it is one of: SEVEN, EIGHT, THREE

Only one of those includes three of the letters that occur at least three times from above, so that line became THREE

Since H only occurs twice, I ran the THREE diagonally up and from the left.

.....E
....E.
...R..
..H...
.T....

First thought would be to place the EIGHT, as that shares the only other H, but that doesn't fit. H is on a single letter space, so let's leave that for now.

Looking at the 2 and 2 3s in the upper right corner, it seems like there should be quite a bit of overlapping there. Let's try placing the NINE and TEN as follows:

..NINE
....E.
...RT.
..H...
.T....

Not bad, could be promising.

So far, we have the following for letter counts

..1122
....2.
...11.
..1...
.1....

Uh-oh! We only have one end N left and it's on SEVEN. That won't fit in our open space. Let's try putting SEVEN where TEN is.

..NINE
....E.
...RV.
..H.E.
.T..S.

..1122
....2.
...11.
..1.1.
.1..1.

How about we run the EIGHT in the second column?

.ENINE
.I..E.
.G.RV.
.HH.E.
.T..S.

.11122
.1..2.
.1.11.
.11.1.
.2..1.

Ooh, I see almost all of TEN. Let's finish that.

TENINE
.I..E.
.G.RV.
.HH.E.
.T..S.

122122
.1..2.
.1.11.
.11.1.
.2..1.

Now, we can add TWO off the T from THREE and ONE off the first E

TENINE
.IONE.
.G.RV.
.HH.E.
.TWOS.

122122
.1113.
.1.11.
.11.1.
.3111.

Hey, let's finish off ZERO

TENINE
.IONE.
.G.RV.
.HH.E.
.TWOSZ

122122
.1213.
.1.21.
.11.2.
.31111

And we can run FIVE up the right side and FOUR down the left. Note, FOUR could be either up or down, whatever floats your boat.

TENINE
FIONEV
OG.RVI
UHH.EF
RTWOSZ

122123
112131
11.211
111.21
131111

And finish off with SIX

TENINE
FIONEV
OGXRVI
UHHIEF
RTWOSZ

122123
112131
111211
111121
131112
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  • 1
    $\begingroup$ Impressive. Very, very nice! $\endgroup$
    – Mathias711
    May 28, 2015 at 19:30
  • 1
    $\begingroup$ Did anyone notice that NINE is in there twice? $\endgroup$
    – JLee
    May 29, 2015 at 14:40
  • $\begingroup$ @JLee Huh...I hadn't noticed that. $\endgroup$
    – tfitzger
    May 29, 2015 at 15:01

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