Question

2^591/255 find remainder

Answer 1

59! is divisible by 8. So it can be written as 59! = 8k, where k is any natural number.

$\huge \bold{Now, R [ \frac{2 {}^{59! } }{255} ]}$

$\huge \bold { = R [ \frac{2 {}^{8k} }{255} ]}$

$\huge \bold { = R [ \frac{(2 {}^{8}) {}^{k} }{255} ]}$

$\huge \bold { = R [ \frac{256{}^{k} }{255} ]}$

$\huge \bold { = R [ \frac{(255 + 1){}^{k} }{255} ]}$

$\small\bold { = R [ \frac{1{}^{k} }{255} ](Using \: Binomial \: Theorem)}$

=1 (Answer)