kub0x
  • Member for 5 years, 10 months
  • Last seen this week
4 answers
3 votes
3k views
Show that if $n$ is a positive integer with $r$ distinct odd prime factors, then $2^r \mid \varphi(n)$
4 votes

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
What do the variables in the Negative Binomial Distribution mean?
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
Can kurtosis measure peakedness?
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
Is my solution of elementary number theory question correct?
Accepted answer
4 votes

It's a bit tricky but I got the answer: We start with $0 \equiv \frac{3^{p}-1}{2} \pmod{q}$ $3^{p}-1 = 2qk$ $3^{p} = 2qk + 1$ this meants that $1 \equiv 3^{p} \pmod{q}$ By the latter we have that $...

View answer
1 answers
2 votes
282 views
Prove that $Gl_{n}(Z/pZ)$ forms a group under multiplication
Accepted answer
3 votes

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 $F_p$....

View answer
2 answers
0 votes
639 views
Suppose $n^2 + 1$ is a prime number. Then $n + 1$ is also a prime number.
Accepted answer
3 votes

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'...

View answer
2 answers
1 votes
217 views
Combinatorics on numbers with given digits
Accepted answer
3 votes

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 $3$ ...

View answer
2 answers
0 votes
428 views
every element of A is equal to its co-factor
3 votes

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

View answer
1 answers
7 votes
540 views
GCD and the cycle decomposition of a permutation
2 votes

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 ...

View answer
1 answers
0 votes
28 views
Is it possible for Galois Field Multiplication result beyond the field
Accepted answer
2 votes

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 $F_2$. ...

View answer
1 answers
0 votes
43 views
Factors of a sequence resulting from repeated exponentiation
2 votes

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
Prove : $\forall n,a,b \in \mathbb{Z}, n\mid (a+b)$ does not always imply $n\mid ab$.
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
Crack the code using the clues
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
How to deduce the combinations formula without using the variations/permutations formula?
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
Proof of composite
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
A problem from a math contest about powers of two
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
roots of quadratic polynomials are positive integer numbers
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
Sophie Germain prime
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
Probability that cutting a stick at two points forms a triangle?
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
Determinant of an $n\times n$ matrix
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
How to write $A D A^T x$ as $\sum_{j=1}^p A_j D_{jj} A_j^T x$?
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
How many solutions to the equation $C \equiv M^e \;(\mbox{mod}\;n)$?
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
How to show $Aut(\mathbb{Z}_{p^{n}})\cong\mathbb{Z}_{p^{n}-p^{n-1}}$ ,p is an odd prime,$n\geq 2.$
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
equality of different powers for cyclic permutations
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
Why is the product of both numbers always an integer root?
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
What is the most motivating way to introduce primitive roots?
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
permutation and combinations discrete mathematics
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
Show that if $(a,n)=d>1$ and $k$ is any positive integer, then $a^k\not\equiv 1\pmod{n}.$
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 ...

View answer
1 answers
0 votes
173 views
Find the last $2$ digits of $2016^{123}$
1 votes

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^{...

View answer
5 answers
1 votes
173 views
$n^{41}\equiv n\bmod 55$ by Fermat's little theorem
1 votes

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 $...

View answer
1
2 3 4 5