BigBeachBanana
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Solve for [imath]x[/imath].
[math]\dfrac{(x+6)!}{(x+2)!}=1680[/math]
[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.Solve for [imath]x[/imath].
[math]\dfrac{(x+6)!}{(x+2)!}=1680[/math]
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.
I just want to show my amazement at this method, wow! simple & effectiveReducing 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.
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 just want to show my amazement at this method, wow! simple & effective
I thought my post (#7) shows the only non-negative solution. And since the expression is monotonic any solution would be unique.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.
[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]Solve for [imath]x[/imath].
[math]\dfrac{(x+6)!}{(x+2)!}=1680[/math]
Are you suggesting that x=-11 is a solution to (x+6)!/(x+2)! = 1680. I doubt that![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]
show that every integer greater than 11 can be expressed as a sum of two positive composite numbers.
[math]u^2 = 1681 [/math][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 = 41:[/math][math]41 = x^2 + 9x + 19[/math][math]\sqrt(u^2)=\sqrt{1681}[/math][math]u= +/- \sqrt{1681}[/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]
Checking:[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]
[math]\frac{(2+6)!}{(2+2)!} =[/math][math]\frac{8!}{4!}=[/math][math]\frac{40320}{24}=1680[/math][math]x=2[/math]