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Help With Math Problem.

xilluzionx

Senior member
Given that there are three primes in a particualr factorization of some interger , k. Then what is the maximum number of primes possible in other prime factorizations of k?
 
Originally posted by: xilluzionx
Given that there are three primes in a particualr factorization of some interger , k. Then what is the maximum number of primes possible in other prime factorizations of k?

3?

there is only one prime factorziation of any unique number. i think... i dont see how u can prime factorize differently
 
Originally posted by: dighn
Originally posted by: xilluzionx
Given that there are three primes in a particualr factorization of some interger , k. Then what is the maximum number of primes possible in other prime factorizations of k?

3?

that your lucky number?😀
 
What the.... ???? So we're given that a particular prime factorization of integer K has only 3 primes? So like 30 is 2x3x5? I don't see how you can factorize a integer to get other prime factorizations.....
 
Originally posted by: dighn
Originally posted by: xilluzionx
Given that there are three primes in a particualr factorization of some interger , k. Then what is the maximum number of primes possible in other prime factorizations of k?

3?

there is only one prime factorziation of any unique number. i think... i dont see how u can prime factorize differently

Yes, the prime factorization for an integer is unique. So if the factorization of k is a prime factorization, you only have 1 way to do it.

-Tom

EDIT: So if the 1 way contains 3 primes, that's all you can have
 
prime factorization of an integer is unique up to permutations and units.

ignoring units, and permutations, prime factorization is unique.

--

Given there are ONLY 3 primes (a,b,c) in a factorization of K.

Short proof: assume there are not 3 primes. Obviously there cant be less primes than 3. So that means there must be can only be more. That means some other prime, n (such that n != a, n !=b, n!=c) is also a factor of K. This means n|K=abc -> n|abc. That means, n|a, n|b or n|c. Since a, b and c are prime, then n = a, b or c. And that contradicts the assumption.
 
Originally posted by: DanFungus
I believe the magic answer is 42

Yes, it is indeed.
Anyway in relation to the question at hand, I don't know anything about higher level mathematics but it seems to me that the answer would also be three, how can you get different prime factorizations?
 
Originally posted by: kracka
Originally posted by: DanFungus
I believe the magic answer is 42

Yes, it is indeed.
Anyway in relation to the question at hand, I don't know anything about higher level mathematics but it seems to me that the answer would also be three, how can you get different prime factorizations?

KRACKA DOWN!!!
 
Originally posted by: xilluzionx
Given that there are three primes in a particualr factorization of some interger , k. Then what is the maximum number of primes possible in other prime factorizations of k?

Most people here aren't reading the question right. Look at this:

Given that there are three primes in a particualr factorization of some interger...

It says nothing about that particular factorization being the prime factorization.

10, 10 is a particular factorization of 100, and more importantly,

10, 2, 2, 3 is a particular factorization of 120 with 3 prime factors...


However, the question does go funny when it says: Then what is the maximum number of primes possible in other prime factorizations...

suggesting that the first one they were talking about was a prime factorization.
 
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