How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?

• By factorising 1080 we get = 2*2*2*3*3*3*5
  Only 2,3 or multiple of using 2,3 we can take as HCF. 5 can't be taken as a factor of HCF as only one 5 is there. using 5 the minimum number can be made are 5,15,30,45, & 60
  Break the facors of 1080 like
  1. 5*3*2= 60, 3*2=6, 3*2=6 i.e. 30, 6 & HCF is 6 and 30*6*6=180
  2. 5*3=15, 3*2*2=24, 3 i.e. 15,24 & HCF is 3 & 15*24*3 =1080
  3. 2*2*2*5 = 40 & 3*3*3=27 i.e. 40, 27 & HCF is 1. 27*40*1 = 1080
  4. 5*3*3*3*2 = 270 ,2 ,2 i.e HCF of 270 & 2 is 2 . 2*270*2 = 180
  HCF can be 1,2,3,6. Above are the 4 pairs fulfils the condition.
• 7 years agoHelpfull: Yes(0) No(0)