Question 80948
let a=youngest, b=next, c=oldest, and h=house


a*b*c=36 ... a+b+c=h ... a=b ... a < c ... {{{(a^2)c=36}}}


factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, 6 & 6 ... this leaves 1, 4, and 9 as possibilities for {{{a^2}}} 


1,1 & 36 is least likely ... 2,2 & 9 and 3,3 & 4 seem equally likely as possible age scenarios


as you can see, more information is needed to arrive at a unique solution