All Questions
Tagged with prime-factorization divisibility
130
questions
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40
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Why $xy=N\Rightarrow (((x−1)p−1\text{ mod }N)−1)\text{ mod} X=0$
Experimentally, I found that for $N=xy;\ 3<x<y$ where $x,y$ are prime numbers and for prime numbers, $p>x$ the below expression is always true:
$$\Big(\big((x-1)^{p-1}\text{ mod }N\big) -1\...
1
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4
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98
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When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer
$a,b,m,x$ are positive integers.
For which $x>0$ is $f(x)$ an integer?
$$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$
I been trying to play with it, I changed it to:
$$\frac{b^2m-a\left(b+x\right)}{a+m\...
1
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1
answer
73
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When $f(x) = \frac{ax + b}{a -x + 1}$ is an integer
Given that $a$ and $b$ are positive integers. and $$f(x) = \frac{ax + b}{a -x + 1}$$ is an integer, what integer values can x have?
If I could only somehow move $x$ from numerator to the denominator ...
1
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2
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81
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Roots modulo a prime
Let us consider the following result:
Note that in the ring of polynomials with coefficients $mod p$, the product $x(x-1)(x-2)...(x-p+1)$ has all possible roots, and equals $x^p-x$ by Fermat's Little ...
1
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1
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33
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Why is it that the degree of a subextension of $K(a^{1/p})/K$ must have a degree dividing $p$?
$p$ is a prime
$K$ is of characteristic not $p$
$a∈K$
$a^{1/p}∉K$
$a$ is not a root of unity
"Then if we pick an element $b∈K(a^{1/p})$, $b∉K$, then $K(b)/K$ is a non-trivial subextension, thus of ...
0
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2
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165
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If $P,Q,R$ are natural numbers where $P$ and $Q$ are primes, and $Q$ divides $PR$, which of the following is true? [closed]
If $P,Q,R$ are natural numbers where $P$ and $Q$ are primes, and $Q$ divides $PR$, which of the following is true?
$P\mid\,Q$
$P\mid\,R$
$P\mid\,QR$
$P\mid\,PQ$
I know that the last option is ...
-1
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1
answer
131
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Find largest small prime factor [closed]
Let $\mathcal{Q}=19^5+660^5+1316^5$ . We know that $25$ is a factor of $\mathcal{Q}$ . Find with a proof the largest prime factor of $\mathcal{Q}$ not exceeding of $10,000$.
I found by computer ...
1
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3
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107
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Manipulation of Large Numbers
I encountered a puzzling problem (though I don't remember where) to find the prime factorization of the number
$$7^{100}-1$$
I think there may be some kind of trick or technique that one could use to ...
0
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1
answer
72
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P-adic order of a prime
Let $p$ be a prime number and $x \in \mathbb{Z}_{0}$. We define the $p$-adic order of $x$ (notation: $ord_{p}(x))$ as the greatest exponent $\alpha \in \mathbb{N}$ so that $p^{\alpha}|x$.
Now show ...
2
votes
2
answers
821
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How many Positive Integers divide $72^{100}$?
So I began by finding the prime factorization of $72$, which is: $$ 72 = 2^{3} \times 3^{2}$$ then exponentiating by $100$: $$ 72^{100} = 2^{300} \times 3^{200}$$ and so now the question is asking to ...
3
votes
1
answer
168
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Divisibility Problem: How many switches remain in their initial position?
I have been working on this problem:
There is a set of $1000$ switches, which are ordered in a row so that each switch is given a distinct rank from $1$ to $1000$. For example, the $i$-th switch ...
0
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2
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398
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Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$
Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$.
I am stuck with this proof. I thought at first that I need to use contradiction but that did not work. What do you ...
3
votes
1
answer
64
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About the divisors of totient numbers
Are there infinitely many integers that do not divide any totient number?
My try:
If $a|b$ then $\phi(a)|\phi(b)$, so the main question would be equivalent to asking wether there are infinitely many ...
3
votes
2
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765
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How many positive integers are divisors of at least one of $24^5, 20^6$, and $45^7$?
I have a vague idea of prime factorizing 24, 20, and 45, then distributing the exponent, find how many numbers are divisible by them (factors), and then subtract the overcounts. How do I do this?
1
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3
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78
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If $a^5|b^3$, does $a^2|b$?
I have proven that $a^5|b^3$ implies that $a\mid b$ using prime factorization and showing that every prime factor of $a$ will divide every prime factor of $b$. I have to prove or disprove that the ...