modulo

Prove that (∀n ∈ N)(∀x ∈ Z)[x^2n is congruent to 0 or 1 modulo 4]. and Using the previous problem to prove that −1 + 4x + x^2 + 8x^3 + x^4 = 0 has no integer solutions

There's a programming puzzle at Programming Praxis in which one is to calculate values of the Ackermann function A(m, n). One comment was particularly helpful, you can see it here. The function values of A(m, n) are known to get gigantic for m >= 4. Therefore I'd like to approach it with modulo...

Why can't we have modulo negative number? I have never seen this.

Hi, Could anyone please tell me why it is that the numbers returned from this equation, repeat (but upon the repeat they are 65535-x). Start with x=1 then keep looping this: x=(75*(x+1))-INT ((75*(x+1))/65537)*65537-1 That is: x=MOD(75*(x+1),65537)-1 Upon 32768 iterations of this equation...

A number is given. That number is appended to itself n times (n can be as large as 10^10). I need to find the number thus formed modulo m. Let the number be 12 and n be 4, so number thus formed is 12121212. Let us find this number modulo 11, then the answer is 4. I tried to find the pattern...

Hi, I have the following exercise: Choose n = 5, then [3] \oplus [4] = [3+4] = [7] = [2] END. I don't understand what is the passage that makes $[7] = [2]$ Why the set that contains the numbers that have remainders equal to 7 when divided by 5, is the same (contains the same elements), of the...

Hello All, I am new to this forum and would like to say thanks to anyone who takes the time out of their busy schedules to not only view but hopefully help. I have the following question. a) Find the solution for 6x+5=2x+1 in modulo 7 b) Explain if the equation in part "a" has a solution in...

I am struggling with several questions, hope someone could help me!! 1. Find the number of solutions of x^2 + 3x + 1 = 0 (mod 131) What is the algorithm for this kind of questions?? Because 131 is so large that you cannot try numbers... 2. Show 9^55 = 1 (mod 2783) I know 2783 is 11^2 * 23, but...

All, New here...so bear with me. I was attempting to calculate the modulo of a complex number using Microsoft Mathematics (64-bit), For instance... a+bi mod c+di ...but the program explained "no can do." No prob...can do this with other software. But it got me thinking, is this...

For example: 7x≡1(mod31) In this example, the inverse of 7 is 9. How can we find out that 9? What are the steps that I need to do? If I have a general modulo equation: 5x+1≡2(mod6) What is the fastest way to solve it? My initial thought was: 5x+1≡2(mod6) ⇔5x+1−1≡2−1(mod6)...

x == 5 (mod 17) what is the largest 6 figure integer that satisfies x. My answer through brute force: x = 999996 (mod 17) Is there a more subtle way of finding the answer rather than by adding multiples of 17 to 5.

Hi i HAVE A TABLE: n | f(n) 1 | 6 2 | 7 3 | 6 4 | 7 5 | 6 How can I express this with a mod operator? f(n) = n mod what? Is it possible? Thanks

Suppose I have an odd number of the form \prod p_i ^{\beta_i}\prod q_j ^{\gamma_j} where p_i represents every prime congruent to 1 modulo 4 and q_i is every prime congruent to 3 modulo 4. Now I would like to know how to either calculate or determine (with any algorithm) how many of the divisors...

Hi I was trying to do this problem. But i am getting nowhere. Demonstrate that for natural number m and n, that the modulo 3^m residue of n can be determined from its last m digits when expressed in base 12. I am not sure what it is asking. Thanks in advance

help me $$\sum_{k=0}^{62} {2014 \choose k} \pmod{2017}\ =\ ...$$

Hi I was wondering if it is allowed to do something like this: if a \equiv b \pmod{9} then 3a \equiv 3b \pmod{27} or a \equiv b \pmod{27} I wanted to use this for the following problem: prove that 3^{n+1} | 2^{3^n} + 1 for every n \in N I wanted to use induction for this problem. And...

Could anyone please explain how the answer to: 77^39960 mod 39961 is answer: 77^16 = 1, Hence 20204 (77^8) Thanks.

Can anyone help with this please? Show that if ab = -1 (modulo 4) for some integers a, b then either a  = -1 (modulo 4) or b = -1 (modulo 4). Show that if the primes (p1; p2; : : : ; pk) are all congruent to -1 modulo 4 then 4 (p1p2 : : : pk) - 1 has a prime factor which is congruent to -1...

Let p be prime, then for all x: x^p = x (mod p) Show that for all x: x^5 = x (mod 10) x^13 = x (mod 65) Can you deduce and prove a generalisation?

Find the last digit number of 2013^2013^2013^...^2013. Where the powers are 2013 times.