Can anyone help me on this or give me a hint?
The function $f(n)=an!+b$,where $a$ and $b$ are positive integers, is defined for all positive integers. If the range of $f$ contains two numbers that differ by 20, what is the least possible value of $f(1)$?
To rephrase the question: we need to find $a$ and $b$ which minimise $f(1) = a+b$, such that for some $n,m$ we have $(a n! + b) - (a m! + b) = 20$. That is, such that $a (n! - m!) = 20$.
Since the given condition has told us nothing about $b$, we can let it be anything convenient; how small can we make $a$?