Fermat's Little Theorem for `a=7` and `p=11`Solution:Fermat's Little Theorem
`a^(p-1)-=1 ("mod "p)` (where `p` is prime, `a` is not divisible by `p`)
Here `a=7, p=11` (Given)
`p=11` is prime. (First condition holds)
`a=7` is not divisible by `p=11`. (Second condition holds)
Now, we find remainder using Modulo method
`a^(p-1) " mod "p`
`7^10" mod "11`
Here `7^10=(7^2)^5`
`=(7^2" mod "11)^5" mod "11`
`=(49" mod "11)^5" mod "11`
`=5^5" mod "11`
Here `5^5=(5^2)^2*5`
`=(((5^2" mod "11)^2" mod "11)*(5" mod "11))" mod "11`
`=(((25" mod "11)^2" mod "11)*5)" mod "11`
`=((3^2" mod "11)*5)" mod "11`
Here `3^2=(3^2)^1`
`=((9" mod "11)*5)" mod "11`
`=(9*5)" mod "11`
`=45" mod "11`
`=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.
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