Just a Simple Remainder problem

We can see that $4^{2} \equiv 4 \pmod{6}$ So $4^{96} \equiv 4^{48} \equiv 4^{24} \equiv 4^{12} \equiv 4^{6} \equiv 4^{3} \pmod{6}$ The question is equivalent to $64 \pmod{6}$ which is just $\boxed{4}$

$4^{96} \equiv 16^{48} \pmod {6} \equiv 4^{48} \pmod {6}$

$\quad \space \equiv 16^{24} \pmod {6} \equiv 4^{24} \pmod {6}$ (and it repeats)

$\quad \space \equiv 16^{12} \pmod {6} \equiv 16^{6} \pmod {6} \equiv 16^{3} \pmod {6}$

$\quad \space \equiv 4^{3} \pmod {6} \equiv 64 \pmod {6} \equiv \boxed {4} \pmod {6}$

we can simply solve this by forming a series as

4 , 16 , 64 , 256 , 1024 , 4096 ...........................................................................................

in each of the case remainder is 4 hence it can be clearly concluded that the remainder is 4.

$4\equiv(-2)\pmod{6}\\ 4^{2}\equiv(4)\equiv(-2)\pmod{6}\\ 4^{4}\equiv(4^{2})\equiv(4)\equiv(-2)\pmod{6}\\ \Rightarrow\ 4^{16}\equiv(-2)\pmod{6}\\ \Rightarrow\ 4^{32}\equiv(-2)\pmod{6}\\ \Rightarrow\ 4^{96}\equiv(-8)\equiv(4)\pmod{6}.$

4/6 remainder =2==> (4^2)/6 remainder=4==> (4^3)/6 remainder=4==> (4^4)/6 remainder=4==> ... now we got the pattern. 2,4,4,4,...,4. only 4/6 that remainder 2.

so,(4^96)/6 remainder = 4

96 divide 6 is 16 so it is finite has no remainder. the only no. left is 4 so my answer is 4. thats how i got it but i know it has no mathemaical basis

4^96=1^96=1 modulo 3. 4^96=0 modulo 2.

Therefore 4^96 is one more than a multiple of three and even.

Considering integers modulo 6, of the integers 0,1,2,3,4,5 only 4 satisfies the conditions therefore 4^96=4 modulo 6

Just we can find 4^6 =4096. Then 4096/6 remainder=4

4^(96)=4^(16) * 4^(6). Now the remainder when divided by 6 is equal to the remainder in case of 4^(16). Remainder in case of 4^(16) is equal to remainder in case of 4^(4) which is again equal to the remainder in case of 4^(2). So, 16 i.e. 4^2=(6*2)+4. So 4 is the answer.