1. 3^302 mod 5 using Fermat's Little Theorem

Solution:
`3^(302) ("mod "5)`



Step-1: Identify conditions :
`p=5` is prime.

`a=3` is not divisible by `p=5`.

So, Apply the theorem
`3^(5-1)-=1 ("mod "5)`

`3^(4)-=1 ("mod "5)`

Step-2: Reduce the exponent `302`

`302-:4=75` with a remainder of `2`

i.e., `302=4xx75+2`

Step-3: Simplify
We have
`3^(4)-=1 ("mod "5)`

Taking power `75` on both sides

`(3^(4))^75-=1^75 ("mod "5)`

`3^((4xx75))-=1 ("mod "5)`

Multiply by `3^2` on both sides

`3^((4xx75))xx3^2-=1xx3^2 ("mod "5)`

`3^((4xx75)+2)-=3^2 ("mod "5)`

`3^(302)-=9 ("mod "5)`

`3^(302)-=4 ("mod "5)`

`:.` The remainder is `4` when you divide `3^(302)` by `5`

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2. Fermat's Little Theorem for `a=3` and `p=5`

Solution:


Here `a=3, p=5` (Given)

`p=5` is prime. (First condition holds)

`a=3` is not divisible by `p=5`. (Second condition holds)

Now, we find remainder using Modulo method

`a^(p-1) " mod "p`

`3^4" mod "5`

Here `3^4=(3^2)^2`

`=(3^2" mod "5)^2" mod "5`

`=(9" mod "5)^2" mod "5`

`=4^2" mod "5`

Here `4^2=(4^2)^1`

`=16" mod "5`

`=1`

`:.` The remainder is `1`

So the values satisfy the Fermat's Little Theorem.