All Questions
8
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Form of the divisors of a number (Prime Factorization). Is this algorithm-based proof correct? [duplicate]
I am trying to proof the following result:
For a number $n$ whose prime number decomposition is $p_1^{\alpha _1} ... p_m^{\alpha _m}$. Every divisor of $n$ has the form $p_1^{\beta _1} ... p_m^{\beta ...
- 1,231
1
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0
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84
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Find the remainder when the $2006! + \dfrac{4012!}{2006!}$ is divided by $4013$
$$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$
Answer: $x=1553.$
Solution: $$2006!+4012!/2006!=x\pmod{4013}$$
$$(2006!)^2 -2006!x+4012!=0\pmod{4013} (*)$$
$$4\cdot (2006!)^2-4\cdot 2006!x+4\cdot 4012!=...
- 11
2
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Let $x,y>1$ be coprime integers and $g>0$ a real number such that $g^x,g^y$ are both integers. Is it true that $g\in\mathbb N$?
Let:
$x, y\ $ be coprime integers greater than $1$
$g \in \mathbb{R}^+$
$g_,^x \ g^y \in \mathbb{N}$
Proposition: $g \in \mathbb{N}$
I have not managed to prove it. Via the fundamental theorem of ...
- 107
1
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4
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126
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$n \in \mathbb{N}$ has at least 73 two-digit divisors. Prove that one of the divisors is 60.
$n \in \mathbb{N}$ has at least 73 two-digit divisors.
I have these questions:
a) How can I prove that one of the two-digit divisors must be number 60?
b) How can I find a natural number that has $\...
user635053
2
votes
2
answers
150
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A proof of the existence of prime numbers right after 'booting-up' the counting numbers?
Not sure if the following argument is circular in nature or breaks down in some other way.
We've defined the counting numbers $n \ge 1$ with the two familiar binary operations of addition and ...
- 11.5k
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2
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231
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How can I show $q^k$ does not divide ${{n}\choose{q}}$ when $q$ is a prime factor of $n$ and $k$ is the largest integer such that $q^k$ divides $n$?
Suppose that $n \in \mathbb{N}$ is composite and has a prime factor $q$. If $k \in \mathbb{Z}$ is the greatest number for which $q^k$ divides $n$, how can I show that $q^k$ does not divide ${{n}\...
- 2,737
1
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1
answer
311
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Least prime factors in a sequence of consecutive integers and CRT
Below is my attempt to define a simple notation (ordered set of distinct prime factors) that can be used with the Chinese Remainder Theorem.
Please let me know if anything that I said is incorrect, ...
- 10.5k
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2
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A New, Possible Proof of the Infinitude of the Primes?
$$1=1$$
$$2=2$$
$$3=3$$
$4=2\cdot2$ At $4$, the first prime number, $2$, is there as a factor. So I say that at the square of $2$, $2$ comes into play as a prime factor. At this point, $2$ is the ...
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