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de_brook
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What will be the numbers of positive factors of product of n distinct prime numbers?
i was able to get 2^n. pls how do i prove this?
i was able to get 2^n. pls how do i prove this?
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you mean there will exist a 1-1 correspondence between power set of a set A with cardinality n and the set of positive factors of the product of n primescsprof2000 said:In any factor you come up with, it will either be divisible by one of the n unique primes, or not. Whether or not your particular factor is divisible by a prime is independent of whether your factor is divisible by another prime; so how many ways can you make a factor if you have to make n yes/no decisions?
Think about it as a prime factor set. You're finding the cardinality of the power set...
csprof2000 said:Yes. Hence the 2^n.
The factors of product of n distinct primes refer to the numbers that can be multiplied together to get the product of n distinct prime numbers. For example, if n=3, the factors of the product of 2, 3, and 5 would be 1, 2, 3, 5, 6, 10, and 30.
To find the factors of the product of n distinct primes, you can use the prime factorization method. First, find the prime factorization of the product. Then, list out all the possible combinations of the prime factors. Finally, multiply these combinations to get the factors.
No, the factors of the product of n distinct primes are always positive numbers. This is because prime numbers are only divisible by 1 and themselves, and any negative number would result in a non-prime factor.
The number of factors of the product of n distinct primes is equal to 2^n, where n is the number of distinct prime factors. In other words, the number of factors increases exponentially as the number of distinct prime factors increases.
Yes, the concept of factors of product of n distinct primes is used in cryptography and security systems. Prime factorization is a vital tool in the creation of secure encryption algorithms and in breaking them.