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When you factor 1680 you get 2^4×3×5×7 Now to write this as a product of four terms that differ by one, you can find the solution almost immediately : 5×(2×3)×7×2^3 = 5×6×7×8. So, x+6 = 8 => x=2
If you are certain that x is a natural number, you can also shorten the process by guessing: (x + 6)(x + 5)(x + 4)(x + 3) = 1680 Choose the arithmetic mean of the four factors for a new variable: t = x + 4.5 And replace all factors by t for a first guess: t * t * t * t = 1680 t^4 = 1680 Take the square root. Since 40^2 = 1600 and 41^2 = 1681, we get t^2 = 41 approximately. Take the square root once again. Since 6^2 = 36 and 7^2 = 49, we get t = 6.5 approximately. Altogether, we can estimate t = x + 4.5 = 6.5. Subtract 4.5 to estimate x = 2. Then do the proof: x + 3 = 5 x + 4 = 6 x + 5 = 7 x + 6 = 8 And the product is 5 * 6 * 7 * 8 = 30 * 7 * 8 = 210 * 8 = 1680.
By inspection, 1680 is a product of 4 consecutive numbers. It is a lot quicker to work out 4th root of 1680 is between 6 and 7. Hence 1680 is = 5x6x7x8 and x=2 (dont have to deal with factorial of negative number as it is not defined)
Square root of 1680 is about 40. 1680/40 = 42. 42 = 6x7, so easy light bulb moment that 40 = 5x8. Good problem. Looks at first waay difficult, then... Satisfying solve.
Students in the modern era do they have such a patience writing all steps. I have not seen it in USA. All they use scientific calculators. I told my kids several times to write all the steps. They always say they don't need it. They don't use properly the equal signs, too
Don't forget number theory! As it's x! and not gamma function, we have natural numbers in play! The prime factors of 1680 show: 1680 = 2^4 * 3 * 5 * 7 and have to be the product of 4 consecutive natural numbers. Easy to see, that this has to be 5 * 6 * 7 * 8 = 40 * 42 = 1680. You could also substitute u = x + 4.5 (the mean value) and write 1680 = (u - 1.5) (u - 0.5) (u + 0.5) (u + 1.5) = (u - 1.5) (u + 1.5) (u - 0.5) (u + 0.5) = (u² - 1.5²) (u² - 0.5²) and with a formula for (x - a²) (x - b²) = (x - (a²+b²)/2)² - ((a²-b²)/2)² = (x - d² - m²)² - (2dm)² with m = (a+b)/2 and d = (a-b)/2... 1680 = (u² - 1.5²) (u² - 0.5²) = (u² - 0.5² - 1²)² - 1² = 1681 - 1² = 41² - 1² = (41 + 1) (41 - 1) = 42 * 40 = 6 * 7 * 8 * 5. Ergo: u² = 41 + 0.25 + 1 = 42.25 u = 6.5 x = 6.5 - 4.5 = 2 For the calculation: a = 1.5 b = 0.5 d = (1.5 - 0.5) / 2 = 0.5 m = (1.5 + 0.5) / 2 = 1 2dm = 2 * 0.5 * 1 = 1
Actually it's even easier with the similar formula for (x - A) (x - B) = (x - M)² - D² with M = (A+B)/2 and D = (A-B)/2. So easy, that I've overlooked that! A = 1.5² = 2.25 B = 0.5² = 0.25 M = 2.5 / 2 = 1.25 D = 2.0 / 2 = 1.00 1680 = (u² - 1.5²) (u² - 0.5²) = (u² - 1.25)² - 1² = 41² - 1 => u = sqrt(42.25) = 6.5 = x + 4.5 => x = 2.
Al llegar a la ecuación de x2+9x-22=0 es muy simple de factorizar y obtener los valores que se buscan; porque los factores buscados son 11 y -2, ya que al multiplicarlos y restarlos, pues es uno positivo y otro negativo, nos da los elementos del trinomio que tenemos... 11 x (-2) = -22 y 11 - 2 = 9 Por lo que no le veo el caso a lo que hizo después de esa ecuacion cuadrática.
1680 est le produit de quatre entiers successifs et la racine cubique de 1680 est sensiblement 6,4 qui est compris entre 6 et 7, donc il est facile de déduire que les quatre entiers sont 5,6,7 et 8 (trente secondes qui dit mieux ?).
Wow! So complicated solution for so easy task... When you know that (x + 6)(x + 5)(x + 4)(x + 3) = 1680 you can just manually verify integer values [2...6). And 1st is our solution.
It has complex solution/s, since the discriminant is less than zero, and the examiner in the starting of the video is asking for the real solution/s... For that reason it is rejected 👍
Quadratic formula *does* yield solutions for that quadratic. ∆ = b² - 4ac = 81 + 88 = 169 = 13² x = ½(-b ± √∆) = ½(-9 ± 13) = ½{4, -22} = {2, -11} x = 2 is valid; x = -11 is not because the factorials in the original equation become factorials of negative integers, which are undefined / ±∞. Fred
Very insightful! The crucial step is probably the successful attempt to make the expressions in brackets equal and then to substitute them. The formula used to determine the (real) solution to the quadratic equations I replaced it with the formula: minus p/half plus/minus root of p/half squared minus q, I can deal with that better. Thanks for the instructive lecture.
We need 4 consecutive numbers that are multiplied to get 1680. Means the numbers are close to the 4th root of 1680 ie square root of 40+ ie 6+ that's simply 6, 7, 5 (as we need 0) and remaining u can see 8... ... Thus, x is 2.... 8!/4! ie (2+6)!/(2+2)!
13:50 This test can be done much easier. We have 8 * 7 * 6 * 5 = 1680 Rather than tediously multiplying 56 * 30, just divide by 8: 7 * 6 * 5 = 1600/8 = (1600 + 80)/8 = 200 + 10 = 210 Then write 7 * 30 = 210 And you are done.
Let's See. The division of the factorial terms leaves the product x+6 * x+5 * x+4 * x+3 Which means I am looking for 4 consecutive numbers that multiply to be this number. It is obvious that this number is divisible by 80 since 16 divides 8 and 80 divides 80 which gives 80*21=1680 Decomposing these 2 factors gives 2^4 *5 *3 *7 Now I have 7 multiplications that I need to reduce to 4 consecutive multiplications. I'll start with the lowest number 2 and keep multiplying it by 2s until it is no longer lowest, which gives 2^2 * 3*4*5*7 Now I multiply the 3, which is the new lowest, by 2 2^1 * 4*5*6*7 Now I multiply the new lowest which is 4, by 2 to get 5*6*7*8 = 1680 Now the highest term in this product must be x+6 = 8 meaning x=2. 8! /4! = 8*7*6*5 = 1680 I find it interesting how many ways there are ro solve this. I wrote my solution before watching the video or checking the comments, and it is astounding how many other ways people came up with.
No necesariamente, generalmente son los positivos los que usamos, pero también pudieran ser negativos, que casi nunca usamos, porque dos números positivos, lógico que al multiplicarlos da positivo; pero dos números negativos, al multiplicarlos, también dan un número positivo.
(x+6)!/(x+2)!=1680 (x+6)(x+5)(x+4)(x+3)=1680 Note that LHS is a product of four successive numbers, and one of them must be 5 or 10 as 10 | 1680. Note that 1680=16×105 =2×8×3×5×7 =5×6×7×8 The least factor (x+3) is 5 --> x=2 As there are four factors, negative factor is also possible: 1680=(-5)×(-6)×(-7)×(-8) The greatest factor is (x+6). Thus x+6=-5 --> x=-11 Another method: Group the factors into two: the first with the last and the other two in the center. Multiplying the factors in each group we get (x²+9x+18)(x²+9x+20)=1680 [(x²+9x+19)-1][x²+9x+19)+1]=1680 (x²+9x+19)²-1=1680 (x²+9x+19)²=1681 =41² x²+9x+19=±41 x²+9x-22=0 --> (x+11)(x-2)=0 x=2 or x=-11 x²+9x+60=0 --> no real valued solution
@@RoderickEtheria: I have not vome across negative factorial. But I think it is possible to have (-n)! I hope that you are familiar with combination to follow my argument. Combination of n objects taken k each time. There is a symbol for that sentence, but I have difficulty of writing the symbol here. Instead I will write it as nCk defined as nCk=n!/[k!(n-k)!] =n(n-1)(n-2)...(n-k+1)/k! There is formula -nCk as -nCk=(-n)(-n-1)(-n-2)...(-n-k+1)/k! =[(-1)^k]n(n-1)(n-2)...(n+k-1)/k! =[(-1)^k](n+k-1)Ck You will find this in Negative Binomial Distribution. Do you think that (-n)!=[(-1)^n]n! =n! if n is even =-n! if n is odd
@nasrullahhusnan2289 I don't think negative factorial would work here either even if negative factorial was possible. Negative factorial, if possible, would have to head towards -1. But, in this case, the denominator would be farther from -1 than the numerator and should create a fraction less than 1.
@@RoderickEtheria: You are right if x=-11 is plugged back to the original equation: (x+6)!/(x+2)!=1680. But it fits to the simplified equation (x+6)(x+5)(x+4)(x+3)=1680 as LHS=(-5)(-6)(-7)-(8) =5×6×7×8 as there are 4 minus =1680 It's surprise me.
Before watching: Alright, so, (x+6)! = (x+6)(x+5)(x+4)(x+3) (x+2)!. Thus, we can rewrite the left hand side as: (x+6)(x+5)(x+4)(x+3) = 1680 Now, we have some options. We know that (x+6)(x+5)(x+4)(x+3) are a set of 4 consecutive numbers. I think that while we could go through and find the factors of the quartic equation we'll wind up with, it may behoove us to simply factorize 1680. 1680 = (10)(168) = (2)(5)(84)(2) = (2)(5)(3)(7)(2)(2)(2) Now, 5 and 7 are only 2 apart. If we can rearrange the remaining 2s and the 3 to make either 6 and 8 or 4 and 6, we're good. Fortunately, we can! 1680 = (5)(7)(2)(2)(2)(2)(3) = (5)(7)(2*3)(2*2*2) = (5)(7)(6)(8) = (5)(6)(7)(8) = (x+3)(x+4)(x+5)(x+6) Thus, X=2 is our answer.
