modular

If x ≅ y (mod n) and q ≅ z (mod n); then (x +q) ≅ (y + z) (mod n):, i know how to prove this, my question is if we wanted to prove it holds for (x + q) ≅ (y + z) (mod 2n): would this be true or false ?, if true how would i show it ? i know it should be simple but im confused.

How to draw the graph of this function: f(x)=|x+1|-|x-1|

How can you show that 3^54321 - 6 is a multiple of 7? I know you would use modular arithmetic (and maybe the Euclidean algorithm?), but I don't know how to go about doing that. Any help would be greatly appreciated!

Modular electric transmission Golden Globe Awards 2016 Live Stream Packers vs Redskins Live Stream

First problem Let n ∈ N and let Z denote the set of equivalence classes for the relation of congruence modulo n. Let n ∈ Z. Define the order of the element [a]n ∈ Z to be the smallest number k such that : k * [a]n := [a]n + [a]n+....+[a]n = [0]n where the number of terms in the sum is k. (a)...

attached 3 photos of questions I couldn't figure out. Would really appreciate some help especially with the proof involving a partition.

how to find out the value of $\frac ab\pmod m$.

I am trying to find the inverse of 314(mod 7). I have no idea what I am doing wrong.. What I do is: 1) find if gcd(314,7) = 1. Yes! 314 = 7*44 + 6 7 = 6*1 + 1 6 = 1*6 + 0 2) Now to find inverse of 314 I do this: 1 = 7-6 1 = 7 - (314 - 7*44) 1 = 7*45 - 314 And inverse of 314 is -1... What am...

Hi, I am currently working on modular arithmetic and recently I have been investigating on the effects of exponentiation on the system.It will be helpful if someone can share some ideas on the following problem. According to the division algorithm: a= nQ+r, where a is the dividend,n is the...

Is there a simple way to prove that ɸ(p) = p - 1 I would appreciate any suggestions

I need help with the following question please. Calculate 7^42 mod 150

1/192 = k mod7 I ran this through Wolfram Alpha and got the answer k=5 I assume that this is because (192 * 5)/7 has a remainder of 1 (is this correct?) Is there a good way of finding the 5 manually? Thankyou.

I had a young person ask me about modular arithmetic on calculators. I have had very little dealings with modular arithmetic and i didn't think it was a function on most calculators. I found this online modular calculator but I am confused about how it works. you enter the mod you want (m)-...

Does anyone know how to code Mathematica to solve modular problems? For example, say I want to solve 4x == 1 mod 5. My best guess was Mod[Solve[4x == 1, x], 5] but that only resulted in the response Mod[1/4, 5]. Thanks! -Dan

Modular Arithmetic in Z: Definition: a(modm) stands for the remainder when a is divided by m. Definition: a(modm) + b(modm) is (a+b)(modm). Definition: a(modm)xb(modm) is axb(modm). Definition: a≡b(modm): a and b leave same remainder when divided by m Theorem: If a≡b(modm) and c≡d(modm) then...

Hi, all Let 1 + 1/2 + 1/3 .... +1/823 = r/823s Without calculating the left hand side, prove that: r ≡ s (mod8233) I'm comfortable with modular arithmetic so that isnt the problem, I just dont really know how to begin.... Thanks!

so this is confusing, with mod arithmetic i see that you only need the remainder as an answer but i don't get how it is solved. like 2 (equivalence sign) 4(mod 3) (3+4)(mod 5)

I wanted to calculate What is the time complexity of this? I'm using a Java program? Can anyone provide an explanation of the time complexity for the following calculations? c= a * b mod n and m = a ^-1 * b mod n I need any suggest. my variable has 256 bits

I wanted to calculate What is the time complexity of this? I'm using a Java program? Can anyone provide an explanation of the time complexity for the following calculations? c= a * b mod n and m = a ^-1 * b mod n I need any suggest.

Hi, I'm having trouble understanding how to evaluate the last digits of a given extremely large number. I know it has something to do with modular math, but I'm struggling with that still. If anyone could explain how to : Find the last let's say 2 digits of 1402^1402. It would be very...