, 2019

Let $n$ be a 7-digit positive integer made up of the digits 0 and 1 only, and that is divisible by 6.

The leftmost digit of $n$ cannot be 0, so must be 1.

Since $n$ is divisible by 6, then $n$ is even, which means that the rightmost digit of $n$ cannot be 1, and so must be 0.

Therefore, $n$ has the form $1\,pqr\,st0$ for some digits $p,q,r,s,t$ each equal to 0 or 1.

$n$ is divisible by 6 exactly when it is divisible by 2 and by 3.

Since the ones digit of $n$ is 0, then it is divisible by 2.

$n$ is divisible by 3 exactly when the sum of its digits is divisible by 3.

The sum of the digits of $n$ is $1+p+q+r+s+t$.

Since each of $p,q,r,s,t$ is 0 or 1, then $1 \leq 1+p+q+r+s+t \leq 6$.

Thus, $n$ is divisible by 3 exactly when $1+p+q+r+s+t$ is equal to 3 or to 6.

That is, $n$ is divisible by 3 exactly when either 2 of $p,q,r,s,t$ are 1s or all 5 of $p,q,r,s,t$ are 1s.

There are 10 ways for 2 of these to be 1s.

These correspond to the pairs $pq, pr, ps, pt, qr, qs, qt, rs, rt, st$.

There is 1 way for all 5 of $p,q,r,s,t$ to be 1s.

Thus, there are $1 + 10 =11$ such 7-digit integers.