3. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
Answer 36.

Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, because otherwise the product will be at least 121. It is sufficient

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to check $p_{1}=2,3,5,7$. For 2, we get the second factor: $3,5,7,11,13,17,19$, $23,29,31,37,41,43$ or 47 - 14 options, for 3 we get: 3, 5, 7, 11, 13, 17, 19, 23, 29, $31-10$ options, for $5-: 3, \quad 5, \quad 7,11,13,17,19,-7$ options and for $7: 3,5, \quad 7,11,13,-5$ options. In total $14+10+7+5=36$ options.