kub0x
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Let $n=p_1\cdots p_r$ then $\varphi(p_i) = p_i-1 = 2k_i$ thus $\varphi(n)=2k_1\cdots 2k_r = 2^r\cdot(k_1\cdots k_r)$ So $\frac{2^r\cdot(k_1\cdots k_r)}{2^r} = (k_1\cdots k_r) \Rightarrow 2^r \mid \... View answer 1 answers 1 votes 75 views Accepted answer 4 votes The negative binomial distribution is used to measure the probability of k failures before r successes occur. In your case$r=$success,$x=$trials,$x-r=$failures and$p=$success prob. So you have$r$... View answer 3 answers 6 votes 1k views 4 votes From Kurtosis definition: The only data values (observed or observable) that contribute to kurtosis in any meaningful way are those outside the region of the peak; i.e., the outliers. Therefore ... View answer 1 answers 3 votes 106 views Accepted answer 4 votes It's a bit tricky but I got the answer: We start with$0 \equiv \frac{3^{p}-1}{2} \pmod{q}3^{p}-1 = 2qk3^{p} = 2qk + 1$this meants that$1 \equiv 3^{p} \pmod{q}$By the latter we have that$...

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Let $p$ be a prime number. Then the ring of $n\times n$ matrices over $F_p$, which is $\mathcal{M_{n\times n}}(F_p)$ has $GL(n,F_p)$ as a subgroup: the general linear group of dimension $n$ over $F_p$....

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In your reasoning for $n=3$ you have that $n^2+1$ isn't prime. Try to fix a $n$ such that $n^2+1$ is prime but $n+1$ is not. For example: $n=14 \Rightarrow 14^2 + 1 = 197$ is prime but $14+1=15$ isn'...

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You are using permutations on your reasoning but without taking into account the repetition, since for a normal permutation all objects have to be different, this is not the case ($3$ repeats $3$ ...

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$AA^T = I \Rightarrow A^T = A^{-1}$ $A^{-1} = \frac{1}{\det{A}}.C^T = C^T$ since $\det{A}=1$ $A.C^T= I \Rightarrow A = (C^T)^T \Rightarrow A= C$

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I have not seen any name for such permutations, but I can give you some results when every cycle order $\vert c_i \vert$ is either $1$ or a prime not equal to any $\vert c_j \vert$ Taking this case, a ...

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As $GF(2^8)$ is a field it satisfies field axioms, meaning that between others, is closed by multiplication. You've said that multiplication is reduced modulo an irreducible polynomial over $F_2$. ...

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Let $$a_i = 2^{x_i}, x_i = p_1^{p_2^{\cdots ^{p_i}}}$$ Then $b_i = a_i - a_{i-1} = a_{i-1} \cdot (2^{(x_i - x_{i-1})} - 1)$ Remove the trivial factor as: $$\frac{b_i}{a_{i-1}} = 2^{(x_i - x_{i-1})}-... View answer 2 answers 0 votes 71 views Accepted answer 2 votes You could divide it into three cases: 1) If n \mid (a+b) but n \nmid a\cdot b then a+b=nk but both a,b don't have n as a factor \gcd(ab,n)=1. 2) If n \nmid (a+b) then a+b \neq nk but ... View answer 2 answers -1 votes 7k views 2 votes From 4th we know that 7,3,8 are not part of the answer. In 5th we discover 0 as solution and cannot be in 3rd position. Comparing 1 and 2 we know that 6 cannot be part of the answer ... View answer 1 answers 0 votes 755 views 2 votes First when speaking about combinations take into account that order doesn't matter, so for example \{1,2,3\} and \{2,1,3\} are equal. A simple approach for defining mathematical formula of ... View answer 3 answers -1 votes 158 views Accepted answer 2 votes Factorise x^2 + 5x + 4 into (x+1)(x+4). For x even note that one factor is odd and other even. For x odd one factor is even and other odd. Thus always is even and composite. View answer 2 answers 7 votes 159 views 2 votes Basically try to find a power of two that yields 2009 digits. This is done with logarithms as following: log_{10}{2^x} = 2009 which yields x=6673.7535 so 2^{6673} has 2009 digits, 2^{6672} ... View answer 2 answers 5 votes 409 views 2 votes Hint: The trivial case where a=b=c=1 gives you the monic polynomial x^2 - 2x + 1 which by Descarte's Rule of signs has two positive roots and by rational root theorem has two roots both x=1. View answer 1 answers 2 votes 179 views 2 votes A) find all possible options for the remainder of x^p on division by 2p+1 for any integer x. If the order of x in 2p+1 is exactly p then: 1\equiv x^p \pmod{2p+1} \iff Ord_{2p+1}(x)=p ... View answer 1 answers 4 votes 552 views Accepted answer 2 votes As you see in the image linked to the answer of "Bill the Lizard": Now, my confusion. If none of the pieces is greater than half the length of the stick, then why is y>1/2 and x>1/2 in ... View answer 1 answers 0 votes 143 views Accepted answer 2 votes As Crostul says the matrix you have is a Vandermonde Matrix. This matrix has a special property for solving the determinants, but first let's define \alpha. \alpha are the values of the second ... View answer 1 answers 0 votes 29 views Accepted answer 1 votes Check the fact that ADA^Tx=(AD)A^Tx. Then, y=\sum_{j=1}^p (A_j D_{j,j})A^T_jx This is a consequence of matrix multiplication being associative, thus AD is the linear combination of the row ... View answer 1 answers 0 votes 34 views Accepted answer 1 votes Let G=Z_n, M \in G and r = Ord_G(M). There are r unique residues thus r possible exponents e for selected M. Note that r \mid \lambda(n) where \lambda(n) is the Charmichael lambda ... View answer 1 answers 1 votes 49 views 1 votes Let G=Z_{p^n}. How many numbers less than p^n are coprime to p^n? Here coprime numbers are those without p in their factorisation. How many multiples of p are under p^n? Alright p^{n-1} ... View answer 1 answers 1 votes 86 views 1 votes Since the Ord(\tau)=60 we must find a permutation composed of n cycles where each cycle cardinality summed up gives 13 (degree), and lcm gives 60 (order). Based on the factorisation of 60 = 2^... View answer 2 answers 1 votes 92 views Accepted answer 1 votes Let a,b \in \mathbb{Z}. Since the set of integers forms a semigroup under multiplication: a,b \in (\mathbb{Z}, *) then a*b \in (\mathbb{Z},*) and b*a \in (\mathbb{Z},*) (Satisfies closure). ... View answer 1 answers 1 votes 64 views 1 votes Take a cyclic group of prime order minus one this is C_{p-1}. Then the cyclic group C_{p-1} is isomorphic to the group of integers mod p. This is C_{p-1} \cong (\mathbb{Z}/\mathbb{pZ})^{x}. ... View answer 1 answers 0 votes 894 views 1 votes There exist well known formulas for repetition and non-repetition for combination and permutation. Then you need to analyze the question and depending in the presented case (ordering, repetitions, ... View answer 3 answers 2 votes 94 views 1 votes If \gcd(a,n) = d > 1 then d \mid a^k, \forall k\in \mathbb{Z^+} and d \mid n. Now let a^k=(d\cdot x)^k and n=d\cdot y then:$$(d\cdot x)^k \equiv b \pmod {d\cdot y} \Rightarrow (d\cdot ...

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There are two ways of simplifying the congruence to get the desired result: Using your approach: $$(2^5\cdot3^2\cdot7)^{123} \equiv (4^2)^{123}\pmod{4\cdot25}$$ Factor $4$ out: 2^{613}\cdot3^{...
Since $n^{10} \equiv 1 \pmod{11}$ Then $n^{10\cdot k} \equiv 1 \pmod{11}$ Thus for $k=4 \Rightarrow n^{40} \equiv 1 \pmod{11}$ then $n \equiv n^{41} \pmod{11}$ (using Fermat Little's). For modulus \$...