Modular arithmatic exponential

[trap101]

determine if the following congruence has an integer solution: 5^x + 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?

 

[Dick]

5 divides 100. So if two numbers are congruent mod 100, they are also congruent mod 5.

 

[pasmith]

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

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

What happens if x = 0?

 

[trap101]

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.

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.

 

[Dick]

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?

 

[trap101]

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.

 

[Dick]

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?

 

[trap101]

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.

 

[Dick]

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.

 

[rcgldr]

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 ?

 

[trap101]

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)...