Harvard University Algebra Problem | Factorial Aptitude Test | How to Pass Admission Exam
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- Опубликовано: 16 сен 2024
- Harvard University Exam Question || Algebra Problem || Aptitude Simplification Test
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that's easy
x!=(2x+1)(x^3-x)=(2x+1)(x-1)(x)(x+1).
we need to check x=0, x=1 to reject it
for x>1
(x-2)!(x-1)x=(2x+1)(x-1)(x)(x+1). // (x-1)(x)
(x-2)!=(2x+1)(x+1)
let n=x-2, so x=n+2
n!=(2n+5)(n+3)=2n^2 + 11n + 15
for n=0 1!=15 is false so n=0 is not a solution / we can divide both sides by n
(n-1)!=2n + 11 + 15/n n must be a divisor of 15 so we have to check 4 values
n=1 => 0!=1=26 ... false
n=3 => 2!=6=22 false
n=5 => 4!=24=2*5+11+15/5=10+11+3=24 true
n=15 14! is much too big.
Given n = x - 2, we have:
n = 5
x = n + 2
x = 5 + 2
x = 7
x! = (2x + 1)(x^3 - x) = (x - 1)x(x + 1)(2x + 1)
=> (x - 2)! = (x + 1)(2x + 1) = 2x^2 + 3x +1 = (x -2)(2x + 7) + 15
=> (x - 3)! = 2x + 7 + 15/(x - 2) ---(1)
=> x - 2 = one of (1, 3, 5, 15) => x - 2 = 5 meets equation (1)
The equation simplifies to (x-2)!=(x+1)(2x+1). Let t = x-2 > t! = (t+3)(2t+5) > t = 5 > x=7.
This exercise is a converge order if we use for x= 1,-1 and i. The solution will be 0. So the only right solution for this equation has to be 0, or 7
since 0! =1 ,0! = 0 is not correct.
n! , 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 , try let x=7 , 7!=(2*7+1)(7^3-7) , x!=15*336 , 15*336=5040 , 7!=5040 ,
by faktoring , 2x^4+x^3-2x^2-x-5040=0 , 2x^4-14x^3 +15x^3-105x^2 +103x^2-721x +720x-5040=0 ,
(x-7)(2x^3+15x^2+103x+720)=0 , x=7 , 2x^3+15x^2+103x+720=0 , --> roots are non-integer, fractional and complex...
---- this solution would not pass the exam, I think... ----
The approach you have chosen is not an ideal approach (both in terms of brevity and in terms of educational value).
This simple problem can be solved by two a much more elegant and concise approaches: (a) using the concepts of function increment, function growth, and rate of function growth, which are taught in any normal high school; (b) by reducing the problem to an inequality. Here I would like to emphasize my subjective opinion that the approach (a) is more elegant than approach (b), but below I give approach (b) because it is more brevity.
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The original equation (n^2-1)*n*(2*n+1)=n!, n>=1 is equivalent to the equation
(n+1)*n*(2*n+1)=(n-2)!, n>=3.
By introducing the notation m=n-2 (which means m>=5), the last equation takes the form
(m+3)*(2*m+5)=m!, m>=5, or in other words,
2*m^2+11*m+15=m!, m>=5. (1)
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You have already obtained equation (1), but you have not indicated the fact that m>=5 (this fact is important, as you can see below).
From m>=5 it follows that (m-2)!>=(5-2)!=3!=6, i.e. (m-2)!>=6, m>=5.
Therefore, we can write:
m!=(m-2)!*(m-1)*m>=6*(m-1)*m=6*m^2-6*m, m>=5,
i.e. we have the inequality m!>=6*m^2-6*m, m>=5.
Taking this inequality into account in (1), we get:
2*m^2+11*m+15>=6*m^2-6*m, m>=5.
From here, we get the quadratic inequality
4*m^2-17*m-15=5.
The solution to this equation is m=5. Therefore, the solution to the original problem is n=m+2=5+2=7.
That's it!
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Best wishes,
Sharif E. Guseynov,
September 09, 2024, Riga, Latvia
Um isn't it js simply 7..?
(x-2)!= (2x+1)(x+1)
y = x -2
x = y +2
y ! = (2y +3)(y+3)
= 2 y^2 +"9y + 6
Sem finalidade pratica
How do you know?
If you want to study math for practical purposes, just go and be an accountant. Don’t shit in the comment section here.
Factorial Aptitude Test: x! = (2x + 1)(x³ - x); x =?
x! = x(x - 1)(x - 2)!, (2x + 1)(x³ - x) = (2x + 1)(x)(x² - 1) = (2x + 1)(x)(x - 1)(x + 1)
x(x - 1)(x - 2)! = (2x + 1)(x)(x - 1)(x + 1), (x - 2)! = (2x + 1)(x + 1); x > 5
x = 6: (x - 2)! = 4! = 24 < (2x + 1)(x + 1) = (13)(7); x > 6
x = 8: (5 - 2)! = 6! = 720 > (2x + 1)(x + 1) = (17)(9); 8 > x > 6
x = 7: (7 - 2)! = 5! = 120 < (2x + 1)(x + 1) = (15)(8); Proved
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
(2x + 1)(x³ - x) = (15)(7)(7² - 1) = (7)(5)(3)(48) = (7)(6)(5)(4)(3)(2)(1) = 7!; Confirmed
Final answer:
x = 7
Stunning 👌 approach