0809.5: Solution
An Extended WalkThrough, with Diagrams
I suggest you print a grid, and fill it in as you read the following solution.
Since there are twelve entries, there will be twelve entry letters. One key task to solving the puzzle will be to determine the twelve entry letters.
There are one fivespace entry, one sixspace entry, five sevenspace entries, three eightspace entries, and two elevenspace entries. The fivespace entry must include a ‘one’, and no plural ‘s’. All other entries will contain three spaces in addition to the number word: a blank, the entry letter, and a plural ‘s’. The sixspace entry will include a threeletter number word, the sevenspace entries will contain fourletter number words, the eightspace entries will contain fiveletter number words, and the elevenspace entries will contain eightletter number words. So, the following lists will be useful:
Threeletters number words: one, two, six, ten
Fourletters: four, five, nine
Fiveletters: three, seven, eight
Eightletters: thirteen, fourteen, eighteen, nineteen
No other number words until fifty have three, four, five or eight letters, and it quickly becomes clear that there will be fewer than fifty of any given letter in the completed grid.
We start with the fiveletter entry, filling in a ‘one’, though we do not know which letter is described. We do know that it can not be an ‘n’, ‘o’, or ‘e’, because writing it would create two of those letters. Thus, the sixletter horizontal entry which contains the described letter and which must contain a threeletter word, can not have an ‘n’, ‘o’, or ‘e’ in its third space. That is, it must have a ‘six’, which we can fill in.
We now know that six of the twelve entry letters are ‘o’, ‘n’, ‘e’, ‘s’, ‘i’, and ‘x’. We need to find six more.
Consider the vertical entry which contains the ‘s’ from ‘six’. That entry will describe the number of ‘s’s in the puzzle. Let’s fill in all the plural ‘s’s for all the entries in the puzzle. We now have twelve ‘s’s in the puzzle.
The vertical entry which describes the number of ‘s’s must contain one of the eightletter number words. Also, its first letter must be the first letter of a fiveletter word which will go in the horizontal entry which shares that first space. So, there must be either ‘thirteen’ or ‘eighteen’ ‘s’s in the puzzle. We already have twelve ‘s’s. The only way to add more ‘s’s is to write ‘seven’ in the eightspace entries. There are only three eightspace entries, so there is no way to get eighteen ‘s’s into the puzzle. So, there must be thirteen ‘s’s in the puzzle. We can fill in the vertical entry which says so. Remember that since we have twelve ‘s’s already in the puzzle, we will have to add one ‘seven’, in one of the eightspace entries, to get our thirteen ‘s’s.
We now know that ten of the twelve entry letters are ‘o’, ‘n’, ‘e’, ‘s’, ‘i’, ‘x’, ‘t’, ‘h’, ‘r’, and ‘v’. (The ‘v’ comes from the ‘seven’ we will add later.) We need to find only two more letters.
The horizontal entry which shares the first ‘t’ in ‘thirteen’ must contain a ‘three’, since that is the only fiveletter word that begins with ‘t’. The vertical entry which contains the ‘r’ from ‘three’ must contain a ‘four’, since that is the only fourletter word that ends with ‘r’.
The twelve entry letters are thus ‘o’, ‘n’, ‘e’, ‘s’, ‘i’, ‘x’, ‘t’, ‘h’, ‘r’, ‘v’, ‘f’ and ‘u’. ‘Eight’ and ‘eighteen’ will not appear in the coompleted grid.
Consider the horizontal entry which contains the ‘three’. The letter it describes is the third letter of fourletter number word, so it must be a ‘u’, or a ‘v’ or an ‘n’. We already have two ‘n’s in the puzzle, and we know we have to add one more for the ‘seven’ we still have to add to make the number of ‘s’s come out right. If we put an ‘n’ as the entry letter after the ‘three’, we would have four ‘n’s. So, it can’t be an ‘n’. The vertical entry which contains the ‘u’ or the ‘v’ must thus either contain a ‘four’ or a ‘five’.
Consider the horizontal entry which contains the entry letter described by ‘four’ or ‘five’. It is one of two remaining eightletter entries. We know we have to place a ‘seven’ in one of those entries. We already have five ‘e’s in the puzzle. So, the ‘seven’ can not go in the horizontal entry. It must go in the rightmost vertical entry in the puzzle. The eightspace horizontal entry which now contains the ‘v’ from ‘seven’ can not contain an ‘eight’ (no ‘g’s in the puzzle) or a ‘seven’ (we already have all the ‘s’), so it must contain a ‘three’, the only remaining fiveletter number word; there will be three ‘v’s in the completed grid.
Let’s make a table of the letters that will appear in the completed grid, and how many times we have used them, so far:
e 
9 
f 
1 
h 
3 
i 
2 
n 
3 
o 
2 
r 
3 
s 
13 
t 
3 
u 
1 
v 
1 
x 
1 
There will be four or five ‘h’s in the completed grid, we have only three in the grid at the moment. Except for the one open elevenspace entry, all the remaining entries that lack number words are sevenspace entries, and so will contain ‘four’, ‘five’, or ‘nine’; no more ‘h’s! So, the open elevenspace entry must contain the only other eightletter number word that can appear in the grid and that has an ‘h’ in it: ‘thirteen’. Furthermore, we already have nine ‘e’s in the grid, we have to add at least one more (its entry letter) and no other entry can contain a number word greater than ‘nine’. So, the open elevenspace entry must describe the ‘e’. Let’s fill in the ‘thirteen’, and the ‘four’ (describing the ‘h’s) and update our table. I’ll add a column for whether or not the letter has been used as an entry letter, yet, and for how many of each given letter will appear in the final puzzle. Notice that we have discovered that there will be precisely three ‘u’s in the completed puzzle.

number in grid now 
is entry letter in grid? 
how many appear in the completed grid? 
e 
11 
Y 
13 
f 
2 


h 
4 
Y 
4 
i 
3 
Y 

n 
3 


o 
3 
Y 

r 
5 


s 
13 
Y 
13 
t 
5 


u 
2 
Y 
3 
v 
1 
Y 
3 
x 
1 
Y 
1 
We need two more ‘e’s out of the three entries which still lack number words, and which will contain ‘four’, ‘five’, and ‘nine’. So, one of those entries will contain a ‘four’, and the other two will contain either ‘five’ or ‘nine’. Thus, there will be five ‘i’s in the completed puzzle, and we can fill in the entry which describes the number of ‘i’s, and update our table.

number in grid now 
is entry letter in grid? 
how many will appear in the completed grid? 
e 
12 
Y 
13 
f 
3 


h 
4 
Y 
4 
i 
4 
Y 
5 
n 
3 


o 
3 
Y 

r 
5 


s 
13 
Y 
13 
t 
5 


u 
2 
Y 
3 
v 
2 
Y 
3 
x 
1 
Y 
1 
We already have five each of t’s and ‘r’s, so they must go in the entries that contain ‘six’ and ‘seven’. The only other ‘t’ we will add to the grid will come when we write down its entry letter. So the ‘t’ must follow the ‘six’, leaving the ‘r’ to follow the seven.

number in grid now 
is entry letter in grid? 
how many appear in the completed grid? 
e 
12 
Y 
13 
f 
3 


h 
4 
Y 
4 
i 
4 
Y 
5 
n 
3 


o 
3 
Y 

r 
6 
Y 
7 
s 
13 
Y 
13 
t 
6 
Y 
6 
u 
2 
Y 
3 
v 
2 
Y 
3 
x 
1 
Y 
1 
Of the remaining two entries which lack number words, one will contain a ‘four’ (which will give us the missing ‘r’) and one must contain a ‘five’, to give us our missing ‘v’. Since we have three ‘o’s already, and will only add one more, the entry that describes the number of ‘o’s must contain a ‘four’. The last entry that is missing a number word will receive the ‘five’. Write them in. All we are missing now are the entry letters for ‘f’ and for ‘n’. Write ‘f’ after the ‘five’ and ‘n’ after the ‘four’, and go do something useful.