this is the way to do: now as the thing r^5 is in the lcm thus at least one no.must have r^5 as its factor. thus let p has r^5 as one of its factors and q can have r^0 or r^1 or r^2 or r^3 or r^4 or r^5 as its factors. similarly q has one of its factors r^5 and p has r^0 or r^1 or r^2 or r^3 or r^4 or r^5 as factors. thus total 9 cases not ten as we wont count the case twice which is p = q = r^5. similarly u can do with t . final answer: this is for cases of (x,y) ordered duplet extend by taking cases for three nos.
If you live each day as if it was your last, someday you'll most certainly be right.
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
1
