A math teacher thought of a positive integer of two digits. A math teacher thought of a positive integer of two
digits. She wants her two intelligent students Hanna
and Charlie determine the exact number thought.
For this, Hanna is privately told how many positive divisors the number has; Charlie, the sum of the digits of the number.
A brief conversation between Hanna and Charlie is transcribed below:


*

*Charlie: I can not determine the number;

*Hanna: Neither can I, but I can tell if it is even or odd;

*Charlie: Now I know what the number is;

*Hanna: You know? Then I also know.


Suppose that students are honest and there is perfect logic in all the conclusions drawn. Determine the number thought by the teacher, justifying your answer.

The number is $80$.  Charlie was told the sum of digits is $8$.  Hanna was told there are $10$ divisors: the possible numbers with $10$ divisors are $48$ and $80$, and both are even, hence Hanna's first statement.  
Charlie's candidates for the number are $8, 17, 26, 35, 44, 53, 62, 71, 80$.
$8$ has $4$ divisors, but so does $15$; $17$ has $2$ divisors, but so does $2$; and so on for all the other candidates except $80$, which has $10$ divisors, and the only other
integer in $\{1,\ldots,99\}$ with $10$ divisors is $48$ which is also even.  Hence Charlie's second statement.
Hanna now knows the number can't be $48$, because if Charlie had been told the sum of digits was $12$ he couldn't decide whether the number was $48$ or $66$: $66$ has $8$ divisors, and all possible numbers with $8$ divisors ($24, 30, 40, 42, 54, 56, 66, 70, 78, 88$) are even.  Hence Hanna's second statement.
EDIT: This was for a version of the problem where the possible numbers were $1$ to $99$.  It is not correct for the version where the possible numbers are $10$ to $99$.  I'll post a solution for that version if time permits.
EDIT: Where the possible numbers are $10$ to $99$, the number is not $80$, because after Hanna's first statement Charlie would not know that the number is not $17$ (all the possible numbers with $2$ divisors are odd).  
Instead, the number in this case is $30$.  Charlie was told the sum of digits is $3$.  Hanna was told there are $8$ divisors; all possible numbers with $8$ divisors are even, hence Hanna's first statement.  
Now Charlie's candidates for the number are $12, 21, 30$.  $12$ has $6$ divisors, as does $45$; $21$ has $4$ divisors, as does $15$; so Charlie knows it must be $30$.
Hence Charlie's second statement.  The only sums of digits that would have allowed Charlie to make this second statement (without knowing the answer at the beginning statement) are $2$ (with answer $11$), $3$, $14$ (with answer $59$), and $17$ (with answer $89$).
Now Hanna knows the answer is $30$, which is the only number in the intersection
of $\{24, 30, 40, 42, 54, 56, 66, 70, 78, 88\}$ and $\{11,30,59,89\}$.  Hence Hanna's second statement.
How did I come up with the answer?  A process of elimination.  After 
Charlie's first statement the possible numbers are $11,12,\ldots,98$.
After Hanna's first the possibilities are restricted to those with 
$2$, $3$, $8$, $10$, or $12$ divisors ($40$ possibilities).
After Charlie's second statement the possibilities are restricted to
$11, 30, 59, 89$.  And of those all but $30$ have $2$ divisors.
