5^119 mod 59 using Fermat's Little Theorem

Solution:
`5^(119) ("mod "59)`



Step-1: Identify conditions :
`p=59` is prime.

`a=5` is not divisible by `p=59`.

So, Apply the theorem
`5^(59-1)-=1 ("mod "59)`

`5^(58)-=1 ("mod "59)`

Step-2: Reduce the exponent `119`

`119-:58=2` with a remainder of `3`

i.e., `119=58xx2+3`

Step-3: Simplify
We have
`5^(58)-=1 ("mod "59)`

Taking power `2` on both sides

`(5^(58))^2-=1^2 ("mod "59)`

`5^((58xx2))-=1 ("mod "59)`

Multiply by `5^3` on both sides

`5^((58xx2))xx5^3-=1xx5^3 ("mod "59)`

`5^((58xx2)+3)-=5^3 ("mod "59)`

`5^(119)-=125 ("mod "59)`

`5^(119)-=7 ("mod "59)`

`:.` The remainder is `7` when you divide `5^(119)` by `59`

---

---