14) For each integer $n$ between 10 and 99, inclusive, the product of its digits and the sum of its digits are added together, resulting in a new number $S(n)$. For how many $n$ does it happen that $S(n)=n$?
(A) 8,
(B) 9
(C) 10,
(D) 11, (E) 12
14. The answer is (B). If $n$ is an integer between 10 and 99, inclusive, $n$ can be written in the form: $n=10a+b$, where $a$ is the tens digit of $n$ and can vary from 1 to 9, and $b$ is the units digit and can vary from 0 to 9. $n$ has the property required by the exercise if $10a+b=ab+(a+b)$, which means $9a-ab=a(9-b)=0$. Since $a \neq 0$, this means $b-9=0$, i.e., $b=9$. The numbers that have the required property are those that have 9 as the units digit and are between 10 and 99, i.e.: 19, 29, 39, 49, 59, 69, 79, 89, 99; there are 9 in total.