Question

Two natural numbers differ by 3 and their product is 504. Find the numbers.

Answer 1

Let two numbers be x and y

x - y = 3 ------(1)

and xy = 504 ------(2)

then from first x = y + 3 -------(3)

Putting value of eq(3) in eq (2)

(y + 3) y = 504

$y^2+3y-504$ = 0

$y^2+(24-21)y-504$ = 0

$y^2+24y-21y-504$ = 0

y ( y + 24 ) - 21 ( y + 24 ) = 0

( y - 21 ) ( y + 24 )

y = 21 or y = -24

then putting this value of y in eq(1)

we get x = 24 or x = - 21