Least positive rest in division?

jdnhldn
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Homework Statement



Find the least positive rest in division of 7^35 with 5

Homework Equations



(7^35)/5

The Attempt at a Solution



7^35=378818692265664781682717625943 => 378818692265664781682717625943/5... Uhhhhh this is not the way I am supposed to take right? :cry:
 
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What does the phrase "least positive rest" mean? I did a brief google search on it and didn't find anything helpful.
 
By "least positive rest" I think the OP means "remainder." Presumably properties of modular arithmetic should be used to find this remainder.

For example, 7 \equiv 2 (mod 5), so 735 \equiv 235 (mod 5). Does any of this look familiar?
 
jgens said:
What does the phrase "least positive rest" mean? I did a brief google search on it and didn't find anything helpful.
I am sorry that I got you confused by the translation. Yes, as Mark44 mentioned, it means remainder.

Mark44 said:
By "least positive rest" I think the OP means "remainder." Presumably properties of modular arithmetic should be used to find this remainder.

For example, 7 \equiv 2 (mod 5), so 735 \equiv 235 (mod 5). Does any of this look familiar?
This is exactly what I am looking for and it looks familiar :)

But I don't get how 7^35=2^35 (mod 5) Please explain?
 
jdnhldn said:
I am sorry that I got you confused by the translation. Yes, as Mark44 mentioned, it means remainder.


This is exactly what I am looking for and it looks familiar :)

But I don't get how 7^35=2^35 (mod 5) Please explain?

As I already explained, because 7 (mod 5) \equiv 2 (mod 5), then 735 (mod 5)\equiv 235 (mod 5).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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