Modular arithmatic exponential

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Homework Help Overview

The discussion revolves around determining integer solutions for a modular arithmetic problem involving the congruence 5x + 4 ≡ 5 (mod 100). Participants explore properties of congruences and divisibility, particularly focusing on the implications of 5 dividing 100 and the nature of solutions in modular equations. Additionally, there is a related question about finding natural numbers x and y in the context of the fundamental theorem of arithmetic.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants consider various approaches to solve the congruence, including checking values and exploring properties of congruences. Questions arise about the implications of setting x to specific values, such as 0, and the relationship between different modular expressions. There is also a shift to a new problem regarding the relationship between powers of different primes.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions. Some have identified potential solutions, such as x = 0, while others are exploring whether additional solutions exist. The transition to a new question about natural numbers x and y indicates a broadening of the discussion, with participants seeking clarification on the relationships between the powers of different bases.

Contextual Notes

Participants are grappling with the implications of congruences and divisibility, particularly in the context of modular arithmetic. There is an ongoing exploration of how different primes relate to one another in the context of the fundamental theorem of arithmetic, with some confusion noted regarding the division of powers.

trap101
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determine if the following congruence has an integer solution: 5x + 4 congruent to 5 (mod 100)


The only way I could envision this working is by trying all of the remainder values of 100, but I don't think that is the way to go considering there is so many. I tried to figure out an equivalency for 5 \equiv x (mod 100) but that got me nowhere as well.

What's the simple property I'm probably missing here?
 
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5 divides 100. So if two numbers are congruent mod 100, they are also congruent mod 5.
 
trap101 said:
determine if the following congruence has an integer solution: 5x + 4 congruent to 5 (mod 100)

What's the simple property I'm probably missing here?

What happens if x = 0?
 
Dick said:
5 divides 100. So if two numbers are congruent mod 100, they are also congruent mod 5.


is that because we can consider the negative remainder as well? I ask because, I always ever looked at it as 100 had to divide 5, not the other way around.





pasmith said:
What happens if x = 0?

So I guess then it follows since 5 is congruent to 5 (mod 100) that's the same as 5-5 = 100q where q is any integer.
 
trap101 said:
So I guess then it follows since 5 is congruent to 5 (mod 100) that's the same as 5-5 = 100q where q is any integer.

You just mean 'q is some integer', not 'any'. But sure x=0 is a solution. Now you want to check if there are any more solutions. If (5^x+4)-5 is divisible by 100, then it's also divisible by 5. Can that happen if x>0?
 
Dick said:
You just mean 'q is some integer', not 'any'. But sure x=0 is a solution. Now you want to check if there are any more solutions. If (5^x+4)-5 is divisible by 100, then it's also divisible by 5. Can that happen if x>0?

I was about to throw my hands up and give up because I've been here plowing away at numerous other questions...but thinking about it for a second...since I just showed that (5^x+4)-5 is divisible by 5, I just reduced the amount of values I have to check and doing it mentally...all of my values will end up with a final digit as a 9...i.e the only value that works is 5.


Had another question:

find the smallest natural numbers x and y such that (7^2)x = (5^3)y

so this is in a chapter about the fundamental theorem of arithmetic...using that idea, I know I can write each of those powers out into a series of primes...but the primes don't divide each other in any way.
 
trap101 said:
I was about to throw my hands up and give up because I've been here plowing away at numerous other questions...but thinking about it for a second...since I just showed that (5^x+4)-5 is divisible by 5, I just reduced the amount of values I have to check and doing it mentally...all of my values will end up with a final digit as a 9...i.e the only value that works is 5.


Had another question:

find the smallest natural numbers x and y such that (7^2)x = (5^3)y

so this is in a chapter about the fundamental theorem of arithmetic...using that idea, I know I can write each of those powers out into a series of primes...but the primes don't divide each other in any way.

7^2 divides (7^2)x. So it must also divide (5^3)y. What does that tell you about y?
 
Dick said:
7^2 divides (7^2)x. So it must also divide (5^3)y. What does that tell you about y?


I don't think I follow. How does 7^2 also divide (5^3)y?...I see that 7^2 divides (7^2)x and same with the other, but how they divide each other isn't sinking in.
 
trap101 said:
I don't think I follow. How does 7^2 also divide (5^3)y?...I see that 7^2 divides (7^2)x and same with the other, but how they divide each other isn't sinking in.

(5^3)y and (7^2)x are equal. So, of course, if 7^2 divides one, it must divide the other.
 
  • #10
trap101 said:
I don't think I follow. How does 7^2 also divide (5^3)y?
What's he's asking is what is the smallest value for y so that 7^2 also divides (5^3)y ?
 
  • #11
rcgldr said:
What's he's asking is what is the smallest value for y so that 7^2 also divides (5^3)y ?

but isn't that what I'm trying to figure out initially?

I view it as this: (5^3)y/(7^2) = (7^2)q for some q...

==> y = (7^4)q/(5^3)...
 

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