Prove that there is no positive integer $n$ such that, for $k = 1,2,\ldots,9$, the leftmost digit (in decimal notation) of $(n+k)!$ equals $k$.

Call a rational number [i]short[/i] if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m-$[i]tastic[/i] if there exists a number $c\in \{1,2,3,\ldots ,2017\}$ such that $\dfrac{10^t-1}{c\cdot m}$ is short, and such that $\dfrac{10^k-1}{c\cdot m}$ is not short for any $1\le k<t$. Let $S(m)$ be the set of $m-$tastic numbers. Consider $S(m)$ for $m=1,2,\ldots{}.$ What is the maximum number of elements in $S(m)$?

Vukasin, Dimitrije, Dusan, Stefan and Filip asked their teacher to guess three consecutive positive integers, after these true statements: Vukasin: " The sum of the digits of one number is prime number. The sum of the digits of another of the other two is, an even perfect number.($n$ is perfect if $\sigma\left(n\right)=2n$). The sum of the digits of the third number equals to the number of it's positive divisors". Dimitrije:"Everyone of those three numbers has at most two digits equal to $1$ in their decimal representation". Dusan:"If we add $11$ to exactly one of them, then we have a perfect square of an integer" Stefan:"Everyone of them has exactly one prime divisor less than $10$". Filip:"The three numbers are square free". Professor found the right answer. Which numbers did he mention?