Start with the prime factorization of 2400?Is there a fast way to solve this problem? …
Look at this page. \(\displaystyle L\times W=2400\): think divisors.How many different rectangles have an area of 2400? (sides can only be natural numbers) I'm looking for a quick method to solve this problem (not listing)
Mr Bland, I really do not follow your problem with this?Unless I missed a big announcement, is there still no known way to directly determine the prime factors of a number? Do we still have to test all primes less than or equal to the number's square root? If we don't have a formula for factoring numbers, then there isn't a "fast" way to determine how many ways there are to make a rectangle of a particular area.
Working with paper and pencil, I employ divisibility rules, too. That allows me to skip dividing by some primes. With 2400, I'd start dividing by 2, until the quotient is odd. Then I'd test for divisibility by 3. As soon as that test fails, I'd continue with 5 (were the dividend to end in 0 or 5); otherwise I'd consider 7, and so on. Eventually, the quotient is 1.… Do we still have to test all primes less than or equal to the number's square root? …
How is that determined? Through what function [MATH]f(x)[/MATH] can you get [MATH][2^5, 3, 5^2] = f(2400)[/MATH]?Mr Bland, I really do not follow your problem with this?
\(\displaystyle 2400=24\times 10^2=2^5\times 3\times 5^2\) that seems fast to me.
There is no such function. My I suggest a good course in number theory might help.How is that determined? Through what function [MATH]f(x)[/MATH] can you get [MATH][2^5, 3, 5^2] = f(2400)[/MATH]?
If there is no such function, then the process of finding the combinations of rectangles becomes increasingly convoluted as the target area increases, by merit of the necessity to consider additional potential inputs. As such, there is no "fast" method (in a general sense) to solve this problem.There is no such function. My I suggest a good course in number theory might help.
Factoring is hard (typically), but reasonably easy in this case; counting divisors given the prime factorization is easy, using the method pka used in post #5 without explanation.
Presumably the OP doesn't know that method, so maybe we should explain it, or at least help @Darya discover it, rather than bicker.
So here's the idea: We know that 2400 can be written as [MATH]2^5\times 3\times 5^2[/MATH]. That implies that any divisor can be written as
[MATH]2^a\times 3^b\times 5^c[/MATH], where a is between 0 and 5 (inclusive), b is between 0 and 1, and c is between 0 and 2, respectively.
How many ways can you choose a value for each of a, b, and c?
Then you can think about what that tells us about rectangles. (You'll have to decide whether 2 by 1200 and 1200 by 2 are different rectangles.)
nevermind, thank you once again!Factoring is hard (typically), but reasonably easy in this case; counting divisors given the prime factorization is easy, using the method pka used in post #5 without explanation.
Presumably the OP doesn't know that method, so maybe we should explain it, or at least help @Darya discover it, rather than bicker.
So here's the idea: We know that 2400 can be written as [MATH]2^5\times 3\times 5^2[/MATH]. That implies that any divisor can be written as
[MATH]2^a\times 3^b\times 5^c[/MATH], where a is between 0 and 5 (inclusive), b is between 0 and 1, and c is between 0 and 2, respectively.
How many ways can you choose a value for each of a, b, and c?
Then you can think about what that tells us about rectangles. (You'll have to decide whether 2 by 1200 and 1200 by 2 are different rectangles.)
By "function" do you mean "formula"? Entirely too many people, trying to solve a math problem, immediately want a "formula" that will do it for them!How is that determined? Through what function [MATH]f(x)[/MATH] can you get [MATH][2^5, 3, 5^2] = f(2400)[/MATH]?
Through what function [MATH]f(x)[/MATH] can you get [MATH][198733, 777743] = f(154563199619)[/MATH]?
These must be the same function. If such a function exists, then the original rectangle problem can be solved in polynomial time. If such a function exists, cryptosecurity may be in grave peril.
So you're saying you already knew how many divisors there are, and wanted to know the number of rectangles? Here you seem to be saying you want to list them (pairs of sides), but in the OP you explicitly did not want to list.Thank you for your answer. Maybe I just didn't form a question well enough. What I am curious about is how I can find different pairs of sides when I know how many there are divisors, factorization is obvious.