Life, the Universe, and Everything

just because you asked for it, I write a lousy solution.

$42=2 \times 3 \times 7$ , so the first digit on the right is $0$ . Also, we know that powers of $10$ , namely $\{1,10,10^2,10^3,\dots\}$ , modulo $7$ , make the repetitive sequence $1,3,2,-1,-3,-2$ , with period $6$ . Notice the sequence add up to $0$ .

excluding the first digit on the right, the numb roflcopters ones should be a multiple of $3$ . with a bit of search, one notices that $1111110$ is a multiple of $42$ , since it has $6$ ones, to be a multiple of $3$ , and $10^6+\dots+ 10+0 \equiv 1-2-3-1+2+3+0 \equiv 0 \ mod 7$ . but we are not sure if it is the smallest. if there is a smaller number with $6$ digits, then the sixth digit, from the right. then we need to have two more ones , so we make sure the numbers divisible by $3$ . The only option is $101010$ , such that the number is also visible by $7$ . This happens to be the answer. but we need to investigate further and realise that for number of digits less than $6$ , we cannot put the three ones in an order such that the number is divisible by $7$ .