Re: Evaluate 80t-20s+60r-50q+70p
[sp]I think that the best way to do this is to use some linear algebra. Divide the equations by $pqrst$ and write them in matrix form: $$\begin{bmatrix}0&1&1&1&1 \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1 \\ 1&1&1&1&0 \end{bmatrix} \begin{bmatrix}p^{-1} \\q^{-1} \\r^{-1} \\s^{-1} \\t^{-1} \end{bmatrix} = \frac1{pqrst}\begin{bmatrix}-264 \\ -24 \\ 24 \\ 248 \\ -16 \end{bmatrix}.$$ That $5\times5$ matrix is equal to $5P-I$, where $P$ is the matrix of the projection onto the 1-dimensional subspace spanned by the unit vector $\frac1{\sqrt5}(1,1,1,1,1)$. The inverse of that matrix is $\tfrac54P-I = \dfrac14 \begin{bmatrix}-3&1&1&1&1 \\ 1&-3&1&1&1 \\ 1&1&-3&1&1 \\ 1&1&1&-3&1 \\ 1&1&1&1&-3 \end{bmatrix}.$ It follows that $$\begin{bmatrix}p^{-1} \\q^{-1} \\r^{-1} \\s^{-1} \\t^{-1} \end{bmatrix} = \frac1{4pqrst} \begin{bmatrix}-3&1&1&1&1 \\ 1&-3&1&1&1 \\ 1&1&-3&1&1 \\ 1&1&1&-3&1 \\ 1&1&1&1&-3 \end{bmatrix} \begin{bmatrix}-264 \\ -24 \\ 24 \\ 248 \\ -16 \end{bmatrix} = \frac1{pqrst}\begin{bmatrix}256 \\ 16 \\ -32 \\ -256 \\ 8 \end{bmatrix} = \frac1{pqrst}\begin{bmatrix}2^8 \\ 2^4 \\ -2^5 \\ -2^8 \\ 2^3 \end{bmatrix}.$$ If $\Pi = pqrst$ then $p^{-1} = \dfrac{2^8}{\Pi}$, $q^{-1} = \dfrac{2^4}{\Pi}$, $r^{-1} = -\dfrac{2^5}{\Pi}$, $s^{-1} = -\dfrac{2^8}{\Pi}$, $t^{-1} = \dfrac{2^3}{\Pi}.$ Multiply those five equations together to see that $\dfrac1{\Pi} = \dfrac{2^{28}}{\Pi^5}$, from which $\Pi^4 = 2^{28}$ and so $\Pi = 2^7$.
Thus $p = 1/2$, $q = 8$, $r = -4$, $s = -1/2$, $t = 16.$ If I have done the arithmetic correctly, then $80t-20s+60r-50q+70p = 685.$[/sp]
Edit. The left sides of the original equations each contain four of the five products $pqrs$, $pqrt$, $pqst$, $prst$, $qrst$. When I first read the problem, I saw that the first equation uses the four products that include $p$, and the last equation uses the four products that include $t$. I assumed, without stopping to read everything carefully, that if the first equation was a "$p$" equation and the last equation was a "$t$" equation, then the intermediate equations would be the "$q$", "$r$", "$s$" equations, in that order. But anemone points out to me in a PM that the order of the "$q$" and "$r$" equations is reversed. This means that my values for $q$ and $r$ should be exchanged, and that obviously alters the result of the final calculation.