Let the number of pigs be $n$, then the conditions are $$\left\{  \begin{array}{l} n \equiv 3 \pmod{5} \\ n \equiv 1 \pmod{7} \\ n\equiv 0\pmod{3} \end{array} \right.$$

The standard way to solve such a system of congruent relations is to use the Chinese Remainder Theorem which gives the answer as $$x\equiv 3\times 21 \times 1 + 1\times 15\times 1 \equiv \boxed{78} \pmod{105}$$

Alternatively, given a range is provided (less than $100$), it is possible to guess it out. Firstly, by the first relation, we know $x$ must end with either $3$ or $8$. Then, by the second relation, we know it is necessary to check multiples of $7$ which end with $2$ or $7$. The largest such number less than $100$ is $77$. Finally, we find $77+1=\boxed{78}$ indeed is a multiple of $3$. Hence it is the final answer.