2. Let $A$ be the set of all two-digit positive integers $n$ for which the number obtained by erasing its last digit is a divisor of $n$. How many elements does $A$ have?
Answer: 32
Solution: Let $n=10 a+b$. Since $a$ is a divisor of $n$, we infer $a$ divides $b$. Any number $n$ that ends in 0 is therefore a solution. Assuming $b \neq 0, n$ must be one of the numbers $11,12, \ldots, 19,22,24,26$, $28,33,36,39,44,48,55,66,77,88,99$. This gives a total of 32 positive integers.