34. Let $S=\{1,2,3, \ldots, 20\}$ be the set of all positive integers 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 the digits of $N$.
34. Answer: 36
We first find out which two consecutive numbers from $S$ are not factors of $N$. Clearly 1 is a factor of $N$. Note that if an integer $k$ is not a factor of $N$, then $2 k$ is not a factor of $N$ either. Therefore for $2 \leq k \leq 10$, since $2 k$ is in $S, k$ must be a factor of $N$, for otherwise, there would be at least three numbers from $S$ (the two consecutive numbers including $k$, and $2 k$ ) that are not factors of $N$. Hence $2,3, \ldots$, 10 are factors of $N$. Then it follows that $12=3 \times 4,14=2 \times 7,15=3 \times 5,18=2$ $\times 9,20=4 \times 5$ are also factors of $N$. Consequently, since the two numbers from $S$ that are not factors of $N$ are consecutive, we deduce that 11,13 , and 19 are factors of $N$ as well. Thus we conclude that 16 and 17 are the only two consecutive numbers from $S$ that are not factors of $N$. Hence
$$
N=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13 \times 19=6846840,
$$
so the sum of digits of $N=2 \times(6+8+4)=36$.