centenes of $7^{999999}$ 
What is the value of the hundreds digit of the number $7^{999999}$?

Equivalent to finding the value of $a$ for the congruence $$7^{999999}\equiv a\pmod{1000}$$

Using Carmichael function, $$\lambda(1000)=100\implies 7^{100}\equiv1\pmod{1000}$$
Alternatively,
we have $7^2=49=50-1,7^4=(50-1)^2=50^2-2\cdot50\cdot1+1^2=1+2400$
$7^{20}=(7^4)^5=(1+2400)^5=1+\binom512400+\binom52(2400)^2+\cdots\equiv1\pmod{1000}$
Now , $999999\equiv-1\pmod{100}\equiv-1\pmod{20}$
So in either case,  $7^{999999}\equiv7^{-1}\pmod{1000}$
$$\text{Now, }\frac{1000}7=142+\frac67=142+\frac1{\frac76}=142+\frac1{1+\frac16}$$
So, the previous convergent of $\displaystyle\frac{1000}7$ is $\displaystyle142+\frac1{1+\frac11}=\frac{143}1$
Using convergent property (Theorem $3$ here) of continued fraction, $1000\cdot1-7\cdot143=-1$
$\displaystyle\implies 7^{-1}\equiv143\pmod{1000}$
