Fermat's Little Theorem for `a=3` and `p=5`Solution:Fermat's Little Theorem
`a^(p-1)-=1 ("mod "p)` (where `p` is prime, `a` is not divisible by `p`)
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.
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
