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Home > College Algebra calculators > Fermat's Little Theorem example
7. Fermat's Little Theorem example ( Enter your problem )
( Enter your problem )
  1. Example-1 : `3^302 mod 5`
  2. Example-2 : `3^87 mod 5`
  3. Example-3 : `5^284 mod 7`
  4. Example-4 : `5^119 mod 59`
  5. Example-5 : `3^100,000 mod 53`
  6. Example-5 : `a=3` and `p=5`
  7. Example-5 : `a=3` and `p=7`
  8. Example-5 : `a=7` and `p=11`
  9. Example-5 : `a=7` and `p=13`
 		Other related methods
  1. Chinese Remainder Theorem
  2. Extended Euclidean Algorithm
  3. Euclid's Algorithm
  4. Modular multiplicative inverse
  5. Modulo
  6. Fast modular exponentiation
  7. Fermat's Little Theorem
6. Example-5 : `a=3` and `p=5`
(Previous example)		8. Example-5 : `a=7` and `p=11`
(Next example)

7. Example-5 : `a=3` and `p=7`


Fermat's Little Theorem for `a=3` and `p=7`

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=7` (Given)

`p=7` is prime. (First condition holds)

`a=3` is not divisible by `p=7`. (Second condition holds)

Now, we find remainder using Modulo method

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

`3^6" mod "7`

Here `3^6=(3^2)^3`

`=(3^2" mod "7)^3" mod "7`

`=(9" mod "7)^3" mod "7`

`=2^3" mod "7`

Here `2^3=(2^3)^1`

`=8" mod "7`

`=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 Submit Here


6. Example-5 : `a=3` and `p=5`
(Previous example)		8. Example-5 : `a=7` and `p=11`
(Next example)


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