13
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Each number corresponds to a distinct letter. There are some non-words, such as names of things: people, gods, places, etc. Can you reconstruct the crossword?

crossword

In green (1 to 5): Amusing key glues crazy. (5)

Here is a text version of the table, if anyone needs it:

9 10 3 11 12 9 13 14 11 13 15
13 9 16 11 11 11 13 9 6 7
6 2 13 20 3 10 16 17 9 18
6 12 20 16 16 9 17 9
19 20 18 4 11 3 12 9 16
11 16 9 11 1 16 20 21
10 13 20 11 1 7 6 7 3 7
7 1 20 13 9 13 2 17 21
22 11 1 1 11 13 12 7 3 6
20 19 1 2 3 4 5 6 7 8 3
17 11 19 20 9 9 13 9 11
14 9 11 16 18 10 20 3 20 13
9 10 9 6 23 9 3 9 3 7
14 19 7 11 12 6 11 17
14 13 7 8 3 9 13 11 15 11
7 8 11 3 17 9 3 3 11 13
11 3 13 5 11 21 9 6 11 13
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2 Answers 2

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First of all, I see the word "16 16 9 17 9" near the top right. Words starting with double letters are rare, and the only English word which fits this is LLAMA. That gives us:

  • 9 = A
  • 16 = L
  • 17 = M

Then in the upper middle we have a 3-letter word "16 11 11", which must be either LEE or LOO. There are no 6-letter words beginning with AO and ending with A, so it must be LEE, giving:

  • 11 = E

Now we have "13 9 7 9" near the top right and "9 13 9 11" on the right-hand side, leading me to the conclusion that these are RAMA and ARAE. So:

  • 13 = R

The word "A R 6" in the top left must be ART or ARC or ARK (since E and M are taken). The relative frequency of 6, as well as the words "12 6 E M" and "6 R A 21" in the lower right and "E R A 6 7" in the upper right, suggests that:

  • 6 = T

Then "E R A 6 7" must be ERATO (and "7 6 7" in the middle is OTO), giving:

  • 7 = O

And "12 T E M" in the lower right must be STEM, giving:

  • 12 = S

The combination of "S E 3 S E" at the top and "M A 3 3 E R" at the bottom right makes me quite sure that:

  • 3 = N

Now the combination of "23 E N N 5" at the bottom and "T R O 5" in the middle makes me quite sure that:

  • 5 = Y

The words "R 2 M" in the middle and "10 20 N 20 R "on the right tell me that 2 and 20 are both vowels, so they must be I and U in some order. The word "T 8 A 20 N" on the right, as well as the relative frequencies of 2 and 20, gives:

  • 2 = U
  • 20 = I

At this stage I finally solved the cryptic clue:

Amusing key glues crazy = FUNKY (FUN+KY ~ crazy), so

  • 1 = F
  • 4 = K

Mopping up several in quick succession: "22 E F F E R S O N" is obviously JEFFERSON, "F R E U 19" is obviously FREUD, "E I F F E L T O 8 E R" is obviously EIFFEL TOWER; "10 R I E F" could be either BRIEF or GRIEF, but "T U R I N 10" makes clear which one. So:

  • 8 = W
  • 10 = G
  • 19 = D
  • 22 = J

Now "A G N E S A R 14 E R" must be AGNES ARBER and "D I 18 K E N S" must be DICKENS and "F L I 21" must be FLIP and "A G A T 23 A" must be AGATHA, so:

  • 14 = B
  • 18 = C
  • 21 = P
  • 23 = H

Finally, "15 O C A L P O P" must be VOCAL POP, so:

  • 15 = V

And the final solution is:

A G N E S A R B E R V
R A L E E E R A T O
T U R I N G L M A C
T S I L L A M A
D I C K E N S A L
E L A E F L I P
G R I E F O T O N O
O F I R A R U M P
J E F F E R S O N T
I D F U N K Y T O W N
M E D I A A R A E
B A E L C G I N I R
A G A T H A N A N O
B D O E S T E M
B R O W N A R E V E
O W E N M A N N E R
E N R Y E P A T E R

with the letters appearing in the following numerical order:

F U N K Y T O W A G E S R B V L M C D I P J H (with Q, X, Z not present).

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  • $\begingroup$ You have one letter wrong. And I have updated the bottom rows of the text version as Sconibulus noticed I got them wrong while copying them. :) $\endgroup$ Sep 10, 2016 at 17:03
  • $\begingroup$ @MariaDeleva Sorry, can you give me a hint about which letter? I can't find any hole in my logic so far (unless you've used some non-English words in the crossword). There doesn't seem to be any choice for 13 which will fit both A 13 A E and 13 A M A (except L, which is already taken). $\endgroup$ Sep 10, 2016 at 17:08
  • $\begingroup$ This is the letter, but perhaps you are missing a possibility. ;) $\endgroup$ Sep 10, 2016 at 17:15
  • 1
    $\begingroup$ 13 might be R, for Arae (greek) Rama (Hindu) and err(English) $\endgroup$
    – Sconibulus
    Sep 10, 2016 at 17:20
  • 1
    $\begingroup$ @MariaDeleva OK, I think I found it :-) (I did think it odd that there were so many G's ...) $\endgroup$ Sep 10, 2016 at 17:23
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Complete crossword:

 A   G   N   E   S   A   R   B   E   R       V
 R   A       L   E   E       E   R   A   T   O
 T   U   R   I   N   G       L       M   A   C
     T           S   I       L   L   A   M   A
 D   I   C   K   E   N   S       A           L 
     E   L           A   E       F   L   I   P
 G   R   I   E   F       O   T   O       N   O
 O       F   I   R   A       R   U   M       P
 J   E   F   F   E   R   S   O   N       T
I D F U N K Y T O W N M E D I A A R A E B A E L C G I N I R A G A T H A N A N O B D O E S T E M
B R O W N A R E V E O W E N M A N N E R E N R Y E P A T E R

In the following, all ? have to be the same letter. Starting with the beginning of Rand's answer:

First of all, I see the word 16 16 9 17 9 near the top right. Words starting with double letters are rare, and the only English word which fits this is LLAMA. That gives us:

  • 9 = A
  • 16 = L
  • 17 = M

Then in the upper middle we have a 3-letter word 16 11 11, which must be either LEE or LOO. There are no 6-letter words beginning with AO and ending with A, so it must be LEE, giving:

  • 11 = E

Now we have 13 9 7 9 near the top right and 9 13 9 11 on the right-hand side, leading me to the conclusion that these are RAMA and ARAE. So:

  • 13 = R

The word A R 6 in the top left must be ART or ARC or ARK (since E and M are taken). The relative frequency of 6, as well as the words 12 6 E M and 6 R A 21 in the lower right and E R A 6 7 in the upper right, suggests that:

  • 6 = T

Then E R A 6 7 must be ERATO (and 7 6 7 in the middle is OTO), giving:

  • 7 = O

And 12 T E M in the lower right must be STEM, giving:

  • 12 = S

The combination of S E 3 S E at the top and M A 3 3 E R at the bottom right makes me quite sure that:

  • 3 = N

Now the combination of 23 E N N 5 at the bottom and T R O 5 in the middle makes me quite sure that:

  • 5 = Y

Continuing from there:

MEA?O? looks like MEADOW, so 19=D and 8=W

From there,

MED?A then has to be MEDIA, from where 20=I

This leaves an easy clue in the form of:

This forms EI??ELTOWER, so 1=F to form EIFFELTOWER

Also:

LAFO?NTAINE looks like a misspelling of La Fontaine as La Fountaine, so 2=U (this also fits with the crossing R?M, which makes RUM).

Then:

ARNI?A is easy to fill as ARNICA, so 18=C

From there:

?ROWN crosses ?A??, the most likely letter for 14 is then B (BABB is a surname, and no other letter would fit both).

Continuing:

A?NESARBER can only be Agnes Arber, so 10=G. This also forms TURING and GAUTIER.

Another word:

From DIC?ENS, we have 4=K. This forms FUNK?TOWN in the middle, so 5=Y. This solves the cryptic clue:

Amusing key glues crazy = FUNKY (FUN+KY ~ crazy)

Filling in the last letters:

AGAT?A crosses ?ENNY, so 23=H and we get AGATHA and HENNY.
?EFFERSON looks like it could be nothing but a J, so 22=J.

Delivering the final blow, in the top-right corner we have a word with two different unused letters:

15 OCAL 21 O 21. By elimination of the already used letters, I concluded that this should be VOCALPOP (also forming EVE, TRAP and PATER in the bottom-right corner).

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