47 to the power 95 divided by 99..remainder?

$$47^{95}\div99$$ can be divided into two parts. Now we break the divisor 99 into two parts i.e. 11*9.
Let us find the remainder when $$47^{95}\div11$$ and when $$47^{95}\div9$$
$$Rem(47^{95}\div11) = Rem((44+3)^{95}\div11) = Rem(3^{95}\div11) = Rem(243^{19}\div11) = Rem((242+1)^{95}\div11) = 1.$$
$$Rem(47^{95}\div9) = Rem((45+2)^{95}\div9) = Rem(2^{95}\div9) =2^{2} Rem((2^{3})^{31}\div9) = 2^{2}Rem((9-1)^{31}\div9)$$
$$= 2^{2}\times-1 = -4 =5$$
Hence the remainder is a number which leaves a remainder 1 when divided by 11 and a remainder 5 when divided by 9. The smallest number such that is 23.