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# Euler's Theorem watch

1. How would you use Euler's theorem to compute the remainder of A^n + B^m when divided be C.

I'm guessing the division by C means the answer will be the remainder is congruent to something modulo C?

I don't really know how to work this out. Also as A and B are different it makes it harder - I think lol.

Can you help me please?

Thank you
2. (Original post by claret_n_blue)
How would you use Euler's theorem to compute the remainder of A^n + B^m when divided be C.

I'm guessing the division by C means the answer will be the remainder is congruent to something modulo C?

I don't really know how to work this out. Also as A and B are different it makes it harder - I think lol.

Can you help me please?

Thank you
Do you have specific values in mind?

You could reduce them both separately then add the two results.
3. (Original post by ghostwalker)
Do you have specific values in mind?

You could reduce them both separately then add the two results.
C = 143
A^n = 5^124
B^m = 6^242
4. (Original post by claret_n_blue)
C = 143
A^n = 5^124
B^m = 6^242
You have the necesary conditions to apply Euler's theorem to each of the terms. What's causing the problem?
5. (Original post by ghostwalker)
You have the necesary conditions to apply Euler's theorem to each of the terms. What's causing the problem?
So the fact that I'm adding them doesn't change anything? Do I still use the theorem seperately on all the terms? Also, I don't get what I would have to do to find the remainder. If I wrote the answers a = b mod n, then would the remainder simply be 'b'?
6. (Original post by claret_n_blue)
So the fact that I'm adding them doesn't change anything? Do I still use the theorem seperately on all the terms? Also, I don't get what I would have to do to find the remainder. If I wrote the answers a = b mod n, then would the remainder simply be 'b'?
It's basically modular arithmetic and you are looking for the value of the expression mod(143).

Applying Euler's theorem to each of the terms in turn will reduce them to manageable size which you can then work out manually, then add, and finally the division by 143 to get the remainder.

And your answer would just be "Remainder is..."
7. THis is my answer

Spoiler:
Show

143 = 11 x 13

(5^124+6^242)mod 143 = R

(5^120 x 5^4 )mod 143 = (1 x 5^4) mod 143 = 53
(6^240 x 6^2 )mod 143 = (1 x 6^2) mod 143 = 36

R = 53+36 = 89

8. (Original post by noobynoo)
THis is my answer

Spoiler:
Show

143 = 11 x 13

(5^124+6^242)mod 143 = R

(5^120 x 5^4 )mod 143 = (1 x 5^4) mod 143 = 53
(6^240 x 6^2 )mod 143 = (1 x 6^2) mod 143 = 36

R = 53+36 = 89

Why does 5^120 become 1 and how do you get the answer 53 from 5^4 mod 143?

I'm guessing it's the same for the second one as well.
9. (Original post by claret_n_blue)
Why does 5^120 become 1 and how do you get the answer 53 from 5^4 mod 143?

I'm guessing it's the same for the second one as well.

And since 5 is coprime to 143, then by Euler's theorem

and when you divide it by 143, the remainder is 53.

Edit: You could have added the two terms and then divided by 143 (which is what I was suggesting), or this way dividing by 143, and then adding the two. Either is fine.
10. (Original post by ghostwalker)

And since 5 is coprime to 143, then by Euler's theorem

and when you divide it by 143, the remainder is 53.

Edit: You could have added the two terms and then divided by 143 (which is what I was suggesting), or this way dividing by 143, and then adding the two. Either is fine.
And is the second simply 36 because there is no remainder when you divide 36 by 143?
11. (Original post by claret_n_blue)
And is the second simply 36 because there is no remainder when you divide 36 by 143?
36 is the remainder. It's less than 143 already, so you don't need to try and divide 143 into it.

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