3^87 mod 5 using Fermat's Little Theorem

Solution:
`3^(87) ("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 `87`

`87-:4=21` with a remainder of `3`

i.e., `87=4xx21+3`

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

Taking power `21` on both sides

`(3^(4))^21-=1^21 ("mod "5)`

`3^((4xx21))-=1 ("mod "5)`

Multiply by `3^3` on both sides

`3^((4xx21))xx3^3-=1xx3^3 ("mod "5)`

`3^((4xx21)+3)-=3^3 ("mod "5)`

`3^(87)-=27 ("mod "5)`

`3^(87)-=2 ("mod "5)`

`:.` The remainder is `2` when you divide `3^(87)` by `5`

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