Question on factoring a prime

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#1
'Suppose that the prime factorisation of N is , where and . Prove that the number of factors of N is .'

Last edited by username42530853; 1 month ago
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1 month ago
#2
'Suppose that the prime factorisation of N is , where and . Prove that the number of factors of N is .'

Just think of an n-dimensional box with each prime on one dimension.

If necessary, create a simple example with 2 primes, say 2 and 3 and a couple of low exponents. How do you represent each factor on a grid?
Last edited by mqb2766; 1 month ago
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1 month ago
#3
Can't you just solve it with standard combinatorics? Since the question is just asking how many different combinations of the exponents of the factors there are.
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#4
(Original post by JustSomeGuy:/)
Can't you just solve it with standard combinatorics? Since the question is just asking how many different combinations of the exponents of the factors there are.
Pls explain
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1 month ago
#5
Pls explain
Each factor (p) of N appears a times. What I mean is that p^1, p^2, p^3,..., p^a are all factors of N, as said in the question. We also include p^0 as 1 is obviously a factor of N. This means that in total there are a+1 choices for your power of p, each of which provide a different factor of N. Now we extend this to different factors of N (p_1, p_2, p_3, p_4,..., p_n). Each of which having their own unique a and therefore we can state that the factor p_n will have a_n+1 different powers to choose from. If we wish to find all possible combinations for the factors of N, we just multiply all these different combinations together and hence we end up with our final result, (a_1 +1)(a_2 +1)...(a_n +1).
Sorry if I wasn't very clear, feel free to ask for an explanation for any parts that were poorly put.
Last edited by JustSomeGuy:/; 1 month ago
1
#6
(Original post by JustSomeGuy:/)
Each factor (p) of N appears a times. What I mean is that p^1, p^2, p^3,..., p^a are all factors of N, as said in the question. We also include p^0 as 1 is obviously a factor of N. This means that in total there are a+1 choices for your power of p, each of which provide a different factor of N. Now we extend this to different factors of N (p_1, p_2, p_3, p_4,..., p_n). Each of which having their own unique a and therefore we can state that the factor p_n will have a_n+1 different powers to choose from. If we wish to find all possible combinations for the factors of N, we just multiply all these different combinations together and hence we end up with our final result, (a_1 +1)(a_2 +1)...(a_n +1).
Sorry if I wasn't very clear, feel free to ask for an explanation for any parts that were poorly put.
Thank you for explaining, it's all at lot more clear to me now.
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