$7^{19599}$ (in decimal system) the last three digits are

The last three digits of $7^{19599}$ (in the decimal system) are
11. 143.

It is known that $7^{4}=2401=24 \times 100+1$. Therefore (using the binomial theorem), we have $7^{9999}=7^{4} \cdot\left(7^{4}\right)^{2499}=343 \times$ $(2400+1)^{2499}=343 \times(1+2499 \times 2400)\left(\bmod 10^{3}\right)$. Since $343 \times(1+2499 \times 2400)=343 \times$ $(1+(2500-1) \times 2400) \equiv 343 \times(1-2400)=343 \times 601 \equiv 143\left(\bmod 10^{3}\right)$, the last two digits of the required number are still 143.