Let A be the set of first 101 terms of an A.P., whose first term is...

The set $A$ consists of the first 101 terms of an A.P. with first term $a_1 = 1$ and common difference $d_1 = 5$.

The terms of $A$ are $1, 6, 11, 16, \dots, 1 + (101-1) \times 5 = 501$.

The set $B$ consists of the first 71 terms of an A.P. with first term $b_1 = 9$ and common difference $d_2 = 7$.

The terms of $B$ are $9, 16, 23, 30, \dots, 9 + (71-1) \times 7 = 499$.

The first common term of $A$ and $B$ is $16$.

The common difference of the terms in $A \cap B$ is the least common multiple of $d_1$ and $d_2$, which is $\text{lcm}(5, 7) = 35$.

The general term of $A \cap B$ is given by $T_k = 16 + 35k$, where $k \ge 0$ is an integer.

Since the maximum possible value in $A \cap B$ cannot exceed the maximum value in $B$ (which is $499$), we have:

$16 + 35k \le 499$

$35k \le 483$

$k \le 13.8$

Thus, the possible values for $k$ are $0, 1, 2, \dots, 13$.

We require the terms in $A \cap B$ to be divisible by $3$. Therefore:

$16 + 35k \equiv 0 \pmod 3$

$1 + 2k \equiv 0 \pmod 3$

$2k \equiv 2 \pmod 3$

$k \equiv 1 \pmod 3$

The values of $k$ in the range $0 \le k \le 13$ that satisfy $k \equiv 1 \pmod 3$ are $1, 4, 7, 10, 13$.

There are exactly $5$ such values of $k$, which correspond to $5$ elements in $A \cap B$ that are divisible by $3$.

Answer: $5$