Elementary number theory problem Let $X = \{n \in \mathbb{N}: 6 \times n\,\, \text{does  not  consist  of}   \ 0,1,2,3 \, \text{or} \ 4\}.$ For eg, $93 \in X$ because $6 \times 93=558.$
Could anyone advise me how to prove there exists $2$ natural numbers such that the value of every $n \in X$ consists of at least one of them? Thank you. 

$1*6=6$ and $98*6=588$ also $96*6=576$ therefore we wish to prove that any number contains either a $1$ or a $9$
Suppose a number does not contain a 1 or a 9.
Then 6n has more digits than n (because 1 is not a digit). Therefore the left-most digit must be 5. Since there is no 9. So think of doing the multiplication by hand. You get $48=8\cdot6$ plus the tens digit of another number. which is at most five. So the last two digits are between 50 and 53. Clearly the second digit is not a valid number.
i.e. assume a number $n=(a_1a_2a_3a_4\ldots a_k)_{10}\in X$, and $a_i\ne 1,9$.  Since $a_1\ne 1$, the representation of $6n=(b_1b_2b_3\ldots b_{k+1})_{10}$ must have one more digit ($6\cdot 2=12>10$).  But the supposition states $b_i\ne 1,2,3,4$, so $b_1$ and $b_2$ are at least $5$.  Note that $a_1a_2a_3$, however, is at most $888$, since they cannot be $9$.  Then we have $6(a_1a_2a_3)\le 5322$ and so either $b_1\le 4$ or $b_2\le 3$, which is a contradiction.
