If $x_{1},x_{2},x_{3},x_{4},x_{5}$ are distinct positive odd integers and satisfy $(2005-x_{1})(2005-x_{2})(2005-x_{3})(2005-x_{4})(2005-x_{5})=576$. Then the last digit of $x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+x^2_{5}$
Try: $(2005-x_{1})(2005-x_{2})(2005-x_{3})(2005-x_{4})(2005-x_{5})=576 = 24^2= 1\times 6\times 4\times 6\times 4$.
But when we campare , Then at least one number is even.
Could some help me to solve it , Thanks
We need $$\prod_{r=1}^5(2y_i)=576\iff\prod_{r=1}^5y_i=18$$ where $y_i$ are distinct integers
as $2y_i=2005-x_i\iff2y_i<2005$
$y_i$s can be distinct five from $\{\pm1,\pm3,\pm6\}$