Canada | A Nice Algebra Problem | Math Olympiad
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- čas přidán 19. 08. 2024
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Factor 720 to 2^4 x 3^2 x 5, then find 3 consecutive integers you can form by mulitplying the factors, you get 8 x 9 x 10 = 720. X+1 is therefore 7, so X = 6.
I did the same method, but I think, if the number (720) was taller, our method will be too much hard.
The method in the video is so much stylisher than ours ^^
that helps one to find the answer,but does not meet the requirement of the math logic. it's actually a cubic equation that you need to find for this particular math problem.write down that equation certainly matters in an olympiad exam.
When you reached t³-t=9³-9 it means that you guessed that t=9 (x+3)
(x+4)! / (x+1)! = 720
(x+4)(x+3)(x+2) = 10*9*8
x+4=10, x+3=9, x+2=8
x=6
1. x+1, x+2, x+3, x+4 and then x too - are natural
2. 720 = 10*9*8 - the is no other solutions for three natural numbers n+2, n+1, n
X=6 perché 10!/7! = 7!/7! x 8x9x10 =720
This problem has also two complex solutions: t=(-9+-i*sqr(239))/2, then x=-7.5+-i*sqr(239)/2
No, tere is no complex "solutions" because factorial is defined for non-negative interger numbers - x+1, x+2, x+3, x+4 etc. are NATURAL numbers
Otherwise - what is factorial of 8.5+i*sqr(239)/2 ?
the factorial of non-natural numbers is determined by the gamma function. Therefore, other solutions are possible here
NO. There is no consent.
1. The problem uses the factorial symbol, not the gamma function symbol. And the factorial is only defined for natural numbers and 0.
2. "determined" - this is the keyword. The gamma function is only related to the factorial for natural numbers. 𝚪(n) = (n - 1)! - that is, gamma for a natural number is not equal to the factorial of this number. The gamma function is not a generalization of the factorial to complex numbers.
3. To be sure, I checked the gamma function values for those given in the user-nr9cs8fd5q entry
complex solutions of a quadratic equation. They are in no way solutions to the original problem, because for the negative real part the gamma function takes complex values very close to 0 for both the real and imaginary parts. Therefore, you will not be able to get the value 720 by multiplying 3 such numbers.
4. If, as you claim, other solutions are possible, prove it.
@@grzegorzkondracki4630thanks for explanation
X=6
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