2^1680 divided by 1763 then remainder will be ? plz explain the answer. ?

Question

2^1680 divided by 1763 then remainder will be ? plz explain the answer. ?

Answer 1

Let's factor 1680

2*2*2*2*3*5*7 = 1680

2^1680

= 2^(7*5*3*2*2*2*2)

= 128^(5*3*2*2*2*2)

= 34359738368^(3*2*2*2*2) mod 1763

= 214^(3*2*2*2*2)

= 9800344^(2*2*2*2) mod 1763

= 1590^(2*2*2*2)

= 2528100^(2*2*2) mod 1763

= 1721^(2*2)

= 2961841^2 mod 1763

= 1^2

= 1 mod 1763

= 1

Answer 2

1763 = 41 × 43

there are 40 × 42 = 1680 remainders modulo 1763 which are co-prime with 1763, they form multiplicative group mod 1763

2^n is cyclic subgroup of this group, its order must divide 1680

Answer:

2^1680  = 1 (mod 1763)