Q. The product of three natural number is 24000 and HCF is 10. How many such a triplets exists?

• Let the numbers be 10x,10y and 10z. becoz hcf is 10.
  Given xyz = 24 and x,y,z are co–prime.
  The possible values of x,y,z is (1,2,12),(1,3,8),(1,4,6),(2,3,4).
  only four triplets are possible. they are 10,20,120;10,30,80;10,40,60 and 20,30,40
  
  
• 10 years agoHelpfull: Yes(3) No(1)
• 30 arrangement of all 5 pairs (1,3,8)(1 ,6, 4)(2, 3, 4)(12, 2, 1)(24, 1, 1) as 3!+3!+3!+3!+3!
• 10 years agoHelpfull: Yes(1) No(1)
• the triplet combination is 120,20,10 m guessing play around with 120 and 20 with 10 constant and you will get a few combinations:120,20,10; 60,40,10; 30,80,10; 20,30,40 so at least 4 triplets I may have missed some combo.
• 10 years agoHelpfull: Yes(0) No(1)