5^284 mod 7 using Fermat's Little Theorem

Solution:
`5^(284) ("mod "7)`



Step-1: Identify conditions :
`p=7` is prime.

`a=5` is not divisible by `p=7`.

So, Apply the theorem
`5^(7-1)-=1 ("mod "7)`

`5^(6)-=1 ("mod "7)`

Step-2: Reduce the exponent `284`

`284-:6=47` with a remainder of `2`

i.e., `284=6xx47+2`

Step-3: Simplify
We have
`5^(6)-=1 ("mod "7)`

Taking power `47` on both sides

`(5^(6))^47-=1^47 ("mod "7)`

`5^((6xx47))-=1 ("mod "7)`

Multiply by `5^2` on both sides

`5^((6xx47))xx5^2-=1xx5^2 ("mod "7)`

`5^((6xx47)+2)-=5^2 ("mod "7)`

`5^(284)-=25 ("mod "7)`

`5^(284)-=4 ("mod "7)`

`:.` The remainder is `4` when you divide `5^(284)` by `7`

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