Let $S =\{1,2,3,...,20\}$ be the set of all positive from $1$ to $20$ suppose that $N$ is the smallest positive integer such that exactly eighteen numbers from $S$ are factors of $N$ and the only two numbers from $S$ that are not factors of $N$ are consecutive integers. Find the sum of digits of $N$.
We first find out which two consecutive numbers from $S$ are not factors of $N$. Clearly $1$ is the factor of $N$. If $k$ is not factor of $N$ then $2k$ will also be not he factor of $N$.
How will I solve further please help.
Thanks
Hint: start down the possible pairs of nonfactors. It can't be $\{19,20\}$ because then we know $4$ and $5$ are both factors, so $20$ must be. Now try $\{18,19\}$, which fails (why?). Keep going.