Find the remainder when 4⁹⁶ is divided by 6.

$$\frac{{{4^{96}}}}{6},$$ we can write it in this form
$$\frac{{{{\left( {6 - 2} \right)}^{96}}}}{6}$$
Now, Remainder will depend only the powers of -2. So,
$$\frac{{{{\left( { - 2} \right)}^{96}}}}{6},$$   it is same as
$$\frac{{{{\left( {{{\left[ { - 2} \right]}^4}} \right)}^{24}}}}{6},$$   it is same as
$$\frac{{{{\left( {16} \right)}^{24}}}}{6}$$
Now,
$$\frac{{\left( {16 \times 16 \times 16 \times 16{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$
On dividing individually 16 we always get a remainder 4.
$$\frac{{\left( {4 \times 4 \times 4 \times 4{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$
Hence, Required Remainder = 4
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.