How many 7-digit positive integers are made up of the digits 0 and 1 only, and are divisible by 6 ?
(A) 16
(B) 11
(C) 21
(D) 10
(E) 33
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 st 0 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 $1 \mathrm{~s}$ or all 5 of $p, q, r, s, t$ are $1 \mathrm{~s}$. There are 10 ways for 2 of these to be 1 s.

These correspond to the pairs $p q, p r, p s, p t, q r, q s, q t, r s, r t, s t$.

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

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

ANSWER: (B)