prime factorization?
Thanks!
Thanks!
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Yes, that's essentially the idea- really, it's dividing up the number until you get *only* primes. If you need to prime factorize, just start dividing any way you like, and keep going until all you have are primes. For example, start with 90:vinviper1 wrote:I need to work on my prime factorization skills. Is that simply breaking down the number until you get some primes?
Ian Stewart wrote:Definitely prime factorization! First, it's easiest to see how you can factor if you write both exponential expressions with the same base:
4^17 - 2^28 = (2^2)^17 - 2^28 = 2^34 - 2^28
Both 2^34 and 2^28 are divisible by 2^28, so we can factor that out:
2^34 - 2^28 = 2^28 * ( 2^6 - 1)
Now, the expression in brackets is small enough to calculate:
2^28 * ( 2^6 - 1) = 2^28 * (64 - 1) = 2^28 * 63 = 2^28 * 3^2 * 7
Once you get this far, any question about divisibility is easy to answer. For example, what's the largest prime divisor? It's 7.
Hi.nifoui7 wrote:Ian Stewart wrote:Definitely prime factorization! First, it's easiest to see how you can factor if you write both exponential expressions with the same base:
4^17 - 2^28 = (2^2)^17 - 2^28 = 2^34 - 2^28
Both 2^34 and 2^28 are divisible by 2^28, so we can factor that out:
2^34 - 2^28 = 2^28 * ( 2^6 - 1)
Now, the expression in brackets is small enough to calculate:
2^28 * ( 2^6 - 1) = 2^28 * (64 - 1) = 2^28 * 63 = 2^28 * 3^2 * 7
Once you get this far, any question about divisibility is easy to answer. For example, what's the largest prime divisor? It's 7.
Hi Ian,
I know this post is already quite old, but I've been trying to understand prime factorization and I'm stuck here with your explanation.
I'm lost when you go from this 2^34 - 2^28 to this 2^28 * ( 2^6 - 1)
Can you explain plz?
Thx a lot!
Just to add to the explanation above: the factorization we're doing here is one you probably will have done many times in algebra questions. When you see something like x^5 - x^3, we'd almost always want to factor out the term with the smaller power: x^5 - x^3 = x^3(x^2 - 1). We're using the same type of factorization in the question above - it's just that we have a base of 2 instead of x.nifoui7 wrote: Hi Ian,
I know this post is already quite old, but I've been trying to understand prime factorization and I'm stuck here with your explanation.
I'm lost when you go from this 2^34 - 2^28 to this 2^28 * ( 2^6 - 1)
Can you explain plz?
Thx a lot!