Such numbers can have only digits $2,3,5,7$ and all digits should be different digit $2,5$ can be only in first position 1) numbers $2,3,5,7$ are answer 2)for $2$-digits numbers we have $23,37,53,73$ are answer 3) for $3$-digits numbers we have that if second digit is $3$ then third should be $7$ but we can not choose as first digits any number If second digit is $7$ then first and third should be $7$ which is impossible 4) for $n$-digits $n \geq 4$ we should have that there are $3$-digits partial number, what is impossible

Sol from exam Suppose that we have number which hase more than 2 digits That is easy contradicition for more than or equal to 4 numbers because we dont use any prime number second time and if we have 3 digit number $357$ which isn't satisfying condition Then we have $2,3,5,7,23,37,53,73$

This problem is the same as https://codeforces.com/problemset/problem/1562/b which appeared on August 2021.

RagvaloD wrote: Such numbers can have only digits $2,3,5,7$ and all digits should be different digit $2,5$ can be only in first position 1) numbers $2,3,5,7$ are answer 2)for $2$-digits numbers we have $23,37,53,73$ are answer 3) for $3$-digits numbers we have that if second digit is $3$ then third should be $7$ but we can not choose as first digits any number If second digit is $7$ then first and third should be $7$ which is impossible 4) for $n$-digits $n \geq 4$ we should have that there are $3$-digits partial number, what is impossible Good solution but your English is not perfect