## 6. Different Digits

Determine the smallest three-digit number with different digits that has exactly three divisors.

Result: $\quad 169$
## Solution.

Every natural number, except for the number 1, has at least two natural divisors, and numbers have exactly two natural divisors if and only if they are prime. If a number has at least two different prime divisors (let's say $p$ and $q$), the total number of divisors is at least 4 (the number is divisible by $1, p, q$, and $p q$).

Therefore, for a prime number to have exactly three natural divisors, it must have exactly one prime number in its prime factorization. If that prime number appears three or more times in the prime factorization of a number, the number of natural divisors is greater than 3. Why? Let $p$ be a prime factor that appears at least 3 times in the prime factorization of some number. Then, in addition to the number 1 and the number itself, the number is divisible by $p$ and $p \cdot p$, both of which are different from 1 and the number itself, meaning it has at least 4 divisors. If the number is of the form $p \cdot p$, where $p$ is prime, then it has exactly three natural divisors: $1, p$, and $p \cdot p$.

For the number $p \cdot p$ to be a two-digit number, $p$ must be at least 10, but it must also be a prime number. Now we substitute in sequence: for $p=11, p \cdot p=121$, but this number does not have distinct digits. For $p=13, p \cdot p=169$ and this is the desired number.