Fermat's Little Theorem - Explained in 5 minutes!

Fermat's Little Theorem can help us find the remainder when a very large number is divided by a prime number.

[badvertised]

So cool -- and I love that you've set this up in a way that very naturally prompts a curious learner to ask, "So what other operations work like this in modular math?"

[marcusscience]

One cute result of Fermat’s little theorem is that in any base b, for a prime p coprime to b, the recurring radix expansion of 1/p will always have a period length that is some factor of (p-1). For instance, in decimal, 1/13 has 6 (a factor of 12) recurring digits, and 1/41 has 5 (a factor of 40) recurring digits.

If b is a primitive root modulo p, the period length of 1/p is (p-1) itself, the maximum length by the pigeonhole principle. In that case, 1/p is cyclic. Such is the case for decimal 1/7 (6 recurring digits) and 1/23 (22 recurring digits).

[camwizemusic]

Love this video!!! You did a fantastic job pitching to/convincing viewers on the need for this powerful theorem due to the limitations of the calculator's memory - definitely sold! 😊 I really like how you broke down an advanced number theory topic in such a simple way, clearly stating the conditions of the theorem and what each variable represented, providing intuitive vocabulary for symbols that may be unfamiliar to some viewers, providing an example with small values of a and p to give everyone a chance to understand intuitively what is happening with the remainder, and then finishing with an example that really highlights the practicality and efficiency of the theorem. Thanks for making this video and explaining it in such a great way! Number Theory is such an fascinating field and one of my favorite areas of mathematics!

[ArthurOgawa-qz]

It took me a while to understand what you were doing at minute 3:20. For the benefit of someone with the same lack of penetration as I: The trick is, in the evaluation of 2^{502} (mod 5), to factor out from 2^{502} all those numbers that are congruent to 1 (mod 5), that is, factors of 16 = 2^4.
The thought crosses my mind: can we not as well factor out 2^2 = 4? We have 4 = -1 (mod 5), after all. That would bring us to, in this case, 2^{502} = 16^{125} * 4^{1} = 1^{125} * (-1)^{1} = -1 (mod 5). Arguably a trivial elaboration.
Thanks for these posts! I do not take readily to number theory, but you are helping me to enjoy learning it. ~~~~Arthur Ogawa

[Andalfulfulde]

Am thrilled 🙂🙂🙂.

Am at the right place to learn.

A Calculus student ❤❤🎉🎉 5:18

[shaikmohammedasif]

Hi mam what is the remainder when 7^103 is divided by 17

[Problematica.]

Can you find the last three Numbers of the number 19^97

[chimetimepaprika]

I feel deceived because your channel said calculus and here is number theory. Just kidding. Your video and channel are cool and you do a very nice job. Woooo

[erjolalatifllari]

what about when the power is smaller then the mod?

[WashilaSila]

The remainder of 17 ^200 ÷ 18 is 1, How