# Cyclically 130

• Mar 19th 2013, 01:58 AM
geniusgarvil
Cyclically 130
Consider all quadruples of real numbers (x1,x2,x3,x4), such that x1x2x3x4 and each of the four numbers plus the product of the other three equals 130. If S is the sum of all possible values of x4, what is S,
rounded to the nearest integer
?
• Mar 19th 2013, 01:59 AM
geniusgarvil
Re: Cyclically 130
1,2,3,4 are the subscripts of x
• Mar 19th 2013, 07:53 AM
Soroban
Re: Cyclically 130
Hello, geniusgarvil!

Quote:

Consider all quadruples of real numbers $\displaystyle a,b,c,d$ , such that $\displaystyle a \le b \le c \le d$

and each of the four numbers plus the product of the other three equals 130.

If $\displaystyle S$ is the sum of all possible values of $\displaystyle d$, what is $\displaystyle S$? .(Round to nearest integer.)

We have: .$\displaystyle \begin{Bmatrix}a + bcd \:=\:130 \\ b + acd \:=\:130 \\ c + abd \:=\:130 \\ d + abc \:=\:120 \end{Bmatrix}$

Since the four variables are completely interchangeable,
. . we conclude that: .$\displaystyle a \,=\, b \,=\, c \,=\, d.$

The first equation becomes: .$\displaystyle a + a^3 \:=\:130 \quad\Rightarrow\quad a^3 + a - 130 \:=\:0$
. . The only real root is: .$\displaystyle a \,=\,5.$

Therefore: .$\displaystyle a \,=\,b\,=\,c\,=\,d\,=\,5 \quad\Rightarrow\quad S \,=\,5$
• Mar 19th 2013, 07:00 PM
geniusgarvil
Re: Cyclically 130
I,don't know what's wrong in your solution, but the answer's incorrect.