First word on each volume of a dictionary Suppose you have to write a dictionary of a language with 27 letters (Spanish): 5 vowels and 22 consonants (yes, Ñ is included). Each word of this new language is made of two letters one wowel and one consonant. Find what would be the first letter on each volume, knowing that the order is alphabetical and there will be 4 volumes.
Knowing how many words there will be in total (220) and in each volume (55) is an easy task, finding the first word of each book seems to be not straightforward (at least to me). I want to know if there is an easy method other than writing the dictionary.
 A: You have correctly determined that there are $220$ words consisting of one vowel and one consonant and that there will be $55$ words in each of the four volumes if the number of words in each volume is the same.
Observe that each vowel is the first letter of $22$ words since there are $22$ consonants and that each consonant is the first letter of $5$ words since there are $5$ vowels.  
Clearly, the first volume begins with the word $ab$, so its first letter is $a$.  To determine the first letter of the second volume, observe that the number of words beginning with the first few letters of the alphabet:
\begin{array}{c c}
a & 22\\
b & 5\\
c & 5\\
d & 5\\
e & 22
\end{array} 
Since $22 + 5 + 5 + 5 = 37 < 55 < 59 = 22 + 5 + 5 + 5 + 22$, we may conclude that the first letter of the second volume is $e$.  Moreover, since the last word that ends with $e$ is the $59$th word, we know that the first four words of the second volume are $ev$, $ew$, $ey$, $ez$.  That leaves $51$ words in the second volume.
Can you continue?  
A: Write the alphabet ${\tt a}$-${\tt z}$ in a first column. Begin the second column with $22$ and add $5$ at each consonant, $22$ at each vowel. If done correctly you end up with $220$. These numbers indicate the total number of words from the beginning of the dictionary until the end of the corresponding letter as first letter. Now find the letter-combinations that would be at positions $1$, $56$, $111$, $166$. By coincidence not much counting back will be necessary, and you get $\ {\tt ab}$, $\ {\tt ew}$,$\ {\tt lu}$,$\ {\tt si}$.
There is no "structural" approach to this problem since the data defining the particular instance  not only consist of the numbers $5$ and $22$, but also incorporate a certain "random" $5$-element-subset of $[27]$, which is $1$ of $74\,750$.
