Maths - remainder

Question (1)
16^(8)
=16^(7) * 16
=16^(7) *(14+2)
=14*16^(7) + 2*16^(7)
=14*16^(7) + 2*(14+2)*16^(6)
=14*16^(7) + 2*14*16^(6)+4*16^(6)
=14*16^(7) + 2*14*16^(6) + 4* (14+2)*16^(5)
=14*16^(7) + 2*14*16^(6) + 4*14*16^(5) + 8*16^(5)
.....
=14*16^(7) + 2*14*16^(6) + 4*14*16^(5) + 8*14*16^(4) + 16*14*16^(3)+32*14*16^(2)+ 64*14*16+128*16
=14*16^(7) + 2*14*16^(6) + 4*14*16^(5) + 8*14*16^(4) + 16*14*16^(3)+32*14*16^(2)+ 64*14*16+128*(14+2)
=14*16^(7) + 2*14*16^(6) + 4*14*16^(5) + 8*14*16^(4) + 16*14*16^(3)+32*14*16^(2)+ 64*14*16+128*(14) +256
=14*16^(7) + 2*14*16^(6) + 4*14*16^(5) + 8*14*16^(4) + 16*14*16^(3)+32*14*16^(2)+ 64*14*16+128*(14) + 14*(18) + 4

The remainder is 4.

Question 2
If a 6-digit number A2391B is divisible by both 11 and 9,
then A+2+3+9+1+B must be divisible by 9 ....[1]
and A+3+1- (2+9+B) = 11*integer ...[2]

from [1], A+B+15= 9*integer =18 or 27, i.e. A+B=3 or A+B=12
from [2], A-B-7 = 11*integer = -11 or 0, i.e. A-B=7 or A-B=-4

Solve A+B=3 and A-B=7
2A=10=>.A=5 and B=-2(rejected)
Solve A+B=3 and A-B=-4
2A=-1 => A=-0.5 (rejected)
Solve A+B=12 and A-B=7
2A=19 =>A=9.5 (rejected)
Solve A+B=12 and A-B=-4
2A=8=>A=4 and B=8

The only solution is 423918