Where Quant meets Logic
991 pairs1,(1983)^2 to (991)^2,(993)^2
pairs can be [1,1983], [2,1982].....[1984,1]. {for sqrt {x} and sqrt{y}}but y>x, so, sqrt{y}>sqrt{x}. so, pairs are till before{991,991}hence, there are 990 pairs for root{y} and root{x} . and there fore for x and y too.
sorry there was an error in question kindly try again !
Is the answer 3?
yeah !! it is 3
(1,7) (2,6) (3,5)each number in each pair multiplied by sqrt(31)
yaar, we have to give (x,y) pairs. so there are 6 pairs.3 others are (7,1) ,(6,2) , (5,3). correct me if i am wrong.
no the answer would still be three because x must be less than y
991 pairs
ReplyDelete1,(1983)^2 to (991)^2,(993)^2
pairs can be [1,1983], [2,1982].....[1984,1]. {for sqrt {x} and sqrt{y}}
ReplyDeletebut y>x, so, sqrt{y}>sqrt{x}. so, pairs are till before{991,991}
hence, there are 990 pairs for root{y} and root{x} . and there fore for x and y too.
sorry there was an error in question
ReplyDeletekindly try again !
Is the answer 3?
ReplyDeleteyeah !! it is 3
ReplyDelete(1,7) (2,6) (3,5)
ReplyDeleteeach number in each pair multiplied by sqrt(31)
yaar, we have to give (x,y) pairs. so there are 6 pairs.
ReplyDelete3 others are (7,1) ,(6,2) , (5,3). correct me if i am wrong.
no the answer would still be three because x must be less than y
ReplyDelete