Question 1071835
<br>The answers posted so far both let one of the 6 numbers be x, or 2x.  That leads to solving a quadratic equation that involves an x term.  The algebra is easier if you let the two "middle" numbers be x-1 and x+1; that way, when you add the expressions for the squares of the 6 numbers, the x terms all cancel out, making the solution much easier.<br>
So let the 6 numbers be x-5, x-3, x-1, x+1, x+3, and x+5.  Then the sum of the squares is
{{{(x^2-10x+25)+(x^2-6x+9)+(x^2-2x+1)+(x^2+2x+1)+(x^2+6x+9)+(x^2+10x+25)=6x^2+70}}}
So then
{{{6x^2+70=8284}}}
{{{6x^2=8214}}}
{{{x^2=1369}}}
{{{x=37}}}
And so the 6 numbers are 32, 34, 36, 38, 40, and 42.