The number of terms common to the two A.P.'s 3, 7, 11, ..., 407 and 2, 9, 16, ..., 709 is ___________.

We have two arithmetic progressions.

The first A.P. is $$3,\,7,\,11,\dots ,407$$. Here the first term is $$a_1 = 3$$ and the common difference is $$d_1 = 7 - 3 = 4$$.

Formula for the $$n^{\text{th}}$$ term of an A.P. is $$a_n = a + (n-1)d$$. Putting $$a_n = 407,\; a = 3,\; d = 4$$ we get

$$407 = 3 + (n_1-1)\,4.$$

Subtracting 3 on both sides,

$$404 = (n_1-1)\,4.$$

Dividing by 4,

$$n_1-1 = 101 \;\Longrightarrow\; n_1 = 102.$$

So the first A.P. has 102 terms and its general term can be written as $$T_1 = 3 + 4k,\quad k = 0,1,2,\dots ,101.$$

The second A.P. is $$2,\,9,\,16,\dots ,709$$. Here $$a_2 = 2$$ and $$d_2 = 9 - 2 = 7$$.

Again using $$a_n = a + (n-1)d$$ with $$a_n = 709,\; a = 2,\; d = 7$$,

$$709 = 2 + (n_2-1)\,7.$$

Subtracting 2,

$$707 = (n_2-1)\,7.$$

Dividing by 7,

$$n_2-1 = 101 \;\Longrightarrow\; n_2 = 102.$$

Thus the second A.P. also has 102 terms and its general term is $$T_2 = 2 + 7m,\quad m = 0,1,2,\dots ,101.$$

Any common term must satisfy

$$3 + 4k = 2 + 7m.$$

Re-arranging,

$$4k - 7m = -1.$$

To solve this linear Diophantine equation we work modulo 7. Taking the equation $$4k \equiv -1 \pmod 7$$, we note that $$-1 \equiv 6 \pmod 7$$, so

$$4k \equiv 6 \pmod 7.$$

The multiplicative inverse of 4 modulo 7 is 2 because $$4 \times 2 = 8 \equiv 1 \pmod 7$$. Multiplying by 2 on both sides,

$$k \equiv 2 \times 6 = 12 \equiv 5 \pmod 7.$$

Hence we can write $$k = 5 + 7t,\quad t = 0,1,2,\dots$$.

Substituting this value of $$k$$ back into the expression for the common term,

$$\begin{aligned} T & = 3 + 4k \\[2pt] & = 3 + 4(5 + 7t) \\[2pt] & = 3 + 20 + 28t \\[2pt] & = 23 + 28t. \end{aligned}$$

Therefore the set of all common terms itself forms an A.P. with first term $$23$$ and common difference $$28$$.

Now the largest possible common term cannot exceed the last term of the first A.P., which is $$407$$ (it is already less than the last term 709 of the second A.P.). So we require

$$23 + 28t \le 407.$$

Subtracting 23,

$$28t \le 384.$$

Dividing by 28,

$$t \le \frac{384}{28} = 13.714\ldots$$

Since $$t$$ must be an integer, the greatest allowable value is $$t_{\max} = 13$$. The smallest value is $$t_{\min} = 0$$.

The number of integral values of $$t$$ is therefore

$$t_{\max} - t_{\min} + 1 = 13 - 0 + 1 = 14.$$

Consequently, there are $$14$$ terms common to the two given arithmetic progressions.

So, the answer is $$14$$.