3. The remainder when the number $7^{99}$ is divided by 2550 is
3.343 .

Notice that $2550=50 \times 51$. First, we find the remainders when $7^{*})$ is divided by 5 and 51, respectively.
First, $7^{199}=7 \times 7^{48}=7 \times 49^{19} \equiv 7 \times(-2)^{459}$
$\equiv-14 \times 16^{12} \equiv-14 \times 256^{6} \equiv-14 \times 1^{6}$
$\equiv-14(\bmod 51)$;
Additionally, $7^{(9)}=7 \times 49^{19} \equiv 7 \times(-1)^{49} \equiv-7(\bmod 50)$.
Thus, $\left\{\begin{array}{l}50 x \equiv-14 \times 50, \\ 51 x \equiv-7 \times 51 .\end{array}(\bmod 2550)\right.$

Subtracting the two equations, we get
$$
x \equiv-7 \times 51-(-14) \times 50 \equiv 343(\bmod 2550) \text {. }
$$

Therefore, the remainder when $7^{* x}$ is divided by 2550 is 343.