Algebra Question of The Day - Factorial Edition: Solve (x + 6)! / (x + 2)! = 1680

[BigBeachBanana]

Solve for [imath]x[/imath].
[math]\dfrac{(x+6)!}{(x+2)!}=1680[/math]

 

[blamocur]

Spoiler

x + 4.5 shouldn't be too far from [imath]1680^{1/4}[/imath]

 

[topsquark]

Solve for [imath]x[/imath].
[math]\dfrac{(x+6)!}{(x+2)!}=1680[/math]

(x + 6)! implies that x is an integer greater than -7. If so, there is no solution.

I presume that you mean
[imath]\dfrac{(x+6)(x+5)(x+4)(x+3)(x+2)(x+1)x}{(x+2)(x+1)x} = 1680[/imath]

or
[imath](x + 6)(x + 5)(x + 4)(x + 3) = 1680[/imath]?

-Dan

 

[blamocur]

Spoiler

Assigning [imath]u = x + 4.5[/imath] yields a 4-th degree equation for [imath]u[/imath] which only has even degree members, which means that it can be reduced to a quadratic equation.

 

[Harry_the_cat]

Spoiler

Reducing 1680 to its prime factors, we get:

1680 = 2 x 2 x 2 x 2 x 3 x 5 x 7 = (2 x 2 x 2) x 7 x (2 x 3) x 5 = 8 x 7 x 6 x 5

Now, since 1680 = (x + 6)(x + 5)(x + 4)(x + 3) , it is clear that x=2 is one solution.

 

[topsquark]

Bah! I completely missed the integer solutions. Please disregard my comments above!

-Dan

 

[blamocur]

Spoiler

Assigning [imath]u = x + 4.5[/imath] yields a 4-th degree equation for [imath]u[/imath] which only has even degree members, which means that it can be reduced to a quadratic equation.

Spoiler: Decided to add the details

[imath]u = x+4.5 \Rightarrow (u-1.5)(u-0.5)(u+0.5)(u+1.5) = 1680 \Rightarrow (u^2 - 2.25)(u^2-0.25) = 1680[/imath]
[imath]w = u^2-1.25 \Rightarrow (w-1)(w+1) = 1680 \Leftrightarrow w^2 = 1681 \Rightarrow w = 41 \Rightarrow u^2 = 42.25 \Rightarrow u = 6.5 \Rightarrow x = 2[/imath]

 

[Steven G]

(x+6)(x+5)(x+4)(x+3) = 1680 = 2⁴*3*5*7 = 5*(3*2)(7)(2³). So x+3=5 or x=2

Studying number theory is really helping!

 

[nanase]

Spoiler

Reducing 1680 to its prime factors, we get:

1680 = 2 x 2 x 2 x 2 x 3 x 5 x 7 = (2 x 2 x 2) x 7 x (2 x 3) x 5 = 8 x 7 x 6 x 5

Now, since 1680 = (x + 6)(x + 5)(x + 4)(x + 3) , it is clear that x=2 is one solution.

I just want to show my amazement at this method, wow! simple & effective

 

[Harry_the_cat]

I just want to show my amazement at this method, wow! simple & effective

Thanks. BUT it only gives one solution without confirming that it is the only solution. It happens to be the only solution, but may not have been.

 

[blamocur]

Thanks. BUT it only gives one solution without confirming that it is the only solution. It happens to be the only solution, but may not have been.

I thought my post (#7) shows the only non-negative solution. And since the expression is monotonic any solution would be unique.

 

[pka]

Solve for [imath]x[/imath].
[math]\dfrac{(x+6)!}{(x+2)!}=1680[/math]

[imath]\displaystyle\frac{{\left( {x + 6} \right)!}}{{\left( {x + 2} \right)!}} = \prod\limits_{k = 3}^6 {\left( {x + k} \right)} = {\text{(3 + x) (4 + x) (5 + x) (6 + x) = 1680}}[/imath]
That gives [imath]x^4 + 18 x^3 + 119 x^2 + 342 x + 360=1680[/imath] or SEE HERE

[imath][/imath][imath][/imath][imath][/imath]

 

[Steven G]

[imath]\displaystyle\frac{{\left( {x + 6} \right)!}}{{\left( {x + 2} \right)!}} = \prod\limits_{k = 3}^6 {\left( {x + k} \right)} = {\text{(3 + x) (4 + x) (5 + x) (6 + x) = 1680}}[/imath]
That gives [imath]x^4 + 18 x^3 + 119 x^2 + 342 x + 360=1680[/imath] or SEE HERE

[imath][/imath][imath][/imath][imath][/imath]

Are you suggesting that x=-11 is a solution to (x+6)!/(x+2)! = 1680. I doubt that!

(x+6)!/(x+2)! = (x+6)(x+5)(x+4)(x+3) is NOT always true! There is domain problem here! x>=-2.

 

[Steven G]

@BigBeachBanana,
May I ask what type of calendar that you use? I think that it has been at least one day since you posted your Algebra Question of the Day! I was hoping to see one each day (just not by me).
OK, I ask one. Show that every prime number greater than 11 can be expressed as a sum of two positive composite numbers.

 

[blamocur]

Spoiler

[imath]p = 9 + 2x[/imath] where [imath]x \geq 2[/imath]

 

[Steven G]

How about show that every integer greater than 11 can be expressed as a sum of two positive composite numbers.

 

[blamocur]

show that every integer greater than 11 can be expressed as a sum of two positive composite numbers.

Spoiler

It is [imath]9+2x[/imath] for odd numbers and [imath]4+2x[/imath] for even ones, with [imath]x\geq 2[/imath].

 

[chetheflash]

[math]\frac{(x+6)!}{(x+2)!}=1680 = \frac{(x+6)(x+5)(x+4)(x+3)(x+2)!}{x+2!}[/math][math](x+6)(x+5)(x+4)(x+3)[/math][math](x^2+9x+18)(x^2+9x+20)[/math][math](x^2+9x+19-1)(x^2+9x+19+1)[/math][math](x^2+9x+19)=u[/math]'
[math](u+1)(u-1)=1680[/math][math]{a^2+b^2=(a+b)(a-b)}[/math][math]u^2-1^2=1680=u^2=1680[/math][math]\sqrt(u^2)=\sqrt1680=(-)41[/math][math]u=41/(-41)[/math]:
[math]u=41)[/math]:
[math]x^2+9x+19-41=0[/math][math]x^2+9x-22=0[/math]
[math](b^2-4ac)[/math][math](ax^2+bx+c)[/math][math]9^2-(4x1x(-22)) = 169 > 0[/math][math]u>0[/math][math]u=(-41)[/math][math]x^2+9x+19+41=0[/math][math]x^2+9x+60=0[/math][math]9^2-(4x1x60)[/math][math]81-240=(-159)<0(invalid)[/math][math]x^2+9x-22=x^2+11x-2x-22=0[/math][math]x(x+11)-2(x+11)=x^2+11x-2x-22[/math][math](x+11)(x-2)=0[/math][math]x=(-11)/2[/math][math]x > 0[/math][math]x = 2[/math][math]\frac{(2+6)!}{(2+2)!}=1680[/math][math]\frac{8!}{4!}=1680[/math][math]\frac{40320}{24}=1680[/math][math]x=2[/math]

 

[lookagain]

[math]\frac{(x+6)!}{(x+2)!}=1680 = \frac{(x+6)(x+5)(x+4)(x+3)(x+2)!}{(x+2)!}[/math][math](x+6)(x+5)(x+4)(x+3) = 1680[/math][math](x + 6)(x +3)(x + 5)(x + 4) = 1680 [/math][math](x^2+9x+18)(x^2+9x+20) = 1680 [/math][math](x^2+9x+19-1)(x^2+9x+19+1) = 1680[/math][math]Let \ u = (x^2+9x+19)[/math][math](u-1)(u+1)=1680[/math][math]{a^2-b^2=(a-b)(a+b)}[/math][math]u^2-1^2=1680[/math]

[math]u^2 = 1681 [/math]

[math]\sqrt(u^2)=\sqrt{1681}[/math][math]u= +/- \sqrt{1681}[/math]:
[math]u= +/-41[/math]

[math]u = 41:[/math][math]41 = x^2 + 9x + 19[/math]

[math]x^2+9x+19-41=0[/math][math]x^2+9x-22=0[/math]
[math]The \ discriminant \ is \ (b^2-4ac) \ for [/math][math] \ ax^2+bx+c = 0.[/math][math](9)^2-(4)(1)(-22) = 169 > 0[/math][math]u>0[/math][math]u=-41:[/math]

[math]-41 = x^2 + 9x + 19 [/math]

[math]x^2+9x+19+41=0[/math][math]x^2+9x+60=0[/math][math](9)^2 - (4)(1)(60) = [/math][math]81-240 \ = \ -159 \ <0 \ \ \ (invalid)[/math][math]x^2+9x-22=x^2+11x-2x-22=0[/math][math]x(x+11)-2(x+11) =0[/math][math](x+11)(x-2)=0[/math][math]x= -11, \ 2[/math][math]x > 0[/math][math]x = 2[/math]

Checking:

[math]\frac{(2+6)!}{(2+2)!} =[/math][math]\frac{8!}{4!}=[/math][math]\frac{40320}{24}=1680[/math][math]x=2[/math]

chetheflash, you have many errors, including typos, missing steps, and wrong use of notation, for example. I made suggested corrective changes in the quote box above.

 

[fresh_42]

I was skeptical about whether [imath] 1680 \stackrel{?}{=} (x+6)(x+5)(x+4)(x+3) [/imath] so I used the definition
[math] x!=\Gamma(x+1)=\prod_{n=1}^\infty \left(1+\dfrac{1}{n}\right)^x\left(1+\dfrac{x}{n}\right)^{-1} [/math]and WA to calculate the infinite product
[math] 1680=\dfrac{(x+6)!}{(x+2)!}=\prod_{n=1}^\infty \left(1+\dfrac{1}{n}\right)^{4}\dfrac{n+x+2}{n+x+6} [/math]which resulted in
[math] \begin{gathered} 1680 = (x+6)(x+5)(x+4)(x+3)=x^4 + 18 x^3 + 119 x^2 + 342 x + 360 \\ 0=(x+11)(x-2)(x^2+9x+60) \end{gathered} [/math]Checking the only positive integer solution [imath] x=2 [/imath] yields [imath] 8!/4!=40320/24=1680. [/imath]

It would have been easier if we simply tried [imath] n!/1680 [/imath] for small [imath] n\ge 7[/imath] and see whether [imath] m! [/imath] is the result. Excel would have done it in 2 steps. ;-)