If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?

• Product of HCF and LCM = product of the numbers
  Then, product of the numbers = 19 x 1140
  Let 19a and 19b be the numbers.
  19a x 19b = 19 x 1140
  ab = 19 x 1140 / 19 x 19 = 60
  
  If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
  Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
  Hence the number of such pairs = 3a
• 10 years agoHelpfull: Yes(43) No(21)
• 3 pairs are possible.
  How:
  product of two numbers= 19 x 1140
  both the numbers should be greater than 19. So take two numbers as (19xa=19a) and (19xb=19b)
  19a x 19b = 19x1140 ......(1)
  ab=60.
  If ab=60 then we can have pairs like (1,60) ,(2,30),(3,20),(4,15),(5,12),(6,10).
  Now both the numbers will be prime numbers because if they are not then their HCF will not be 19.
  So possible pair of (a,b) are (1,60),(3,20),(4,15),(5,12).
  Not (1,60) is not allowed because one of the two number will then become 19 and we required both number must be > 19.
  Use (a,b)=(3,20) in Eq. (1)...We got two numbers 57 and 380
  Use (a,b)=(4,15) in Eq. (1)... We got two numbers 76 and 285
  Use (a,b)=(5,12) in Eq. (1)... We got two numbers 95 and 228
  So,at the end we have 3 possible pairs of two numbers which are >19 are (57,380),(76,285),(95,228)
• 9 years agoHelpfull: Yes(20) No(1)
• Let nos be 19a,19b. Then since Hcf=19 and Lcm=1140.
  We get a*b=60. Also given in question,the numbers are greater than 19,(1,60) will not be a pair.Possible pairs for which a*b=60 is (2,30),(3,20),(4,15),(5,12),(6,10) =5
• 10 years agoHelpfull: Yes(14) No(20)
• 19a*19b=19*1140
  a*b=19*1140/361
  3 such pairs are possible
• 10 years agoHelpfull: Yes(11) No(9)
• The numbers should be greater than 19 right?? so how can any pair be the possible answer???
• 10 years agoHelpfull: Yes(9) No(3)
• product of two numbers=hcf*lcm..let two numbers be a & b then a*b=19*1140
  since hcf is 19 number may 19a & 19b where a and b are coprimes.
  so, 19a*19b=19*1140.
  a*b=60.
  so,the possible pairs are (1,60)(2,30)(3,20),(4,15),(5,12),(6,10)..total 6 pairs
  
• 10 years agoHelpfull: Yes(6) No(5)
• since hcf=19
  hence no. are 19a and 19 b.
  product of two numbers= HCF * LCM
  19a *19b = 19 * 1140
  hence ab= 60
  hence pair are (1,60),(2,30), (3,20),(4,15)(5,12),(6,10)
  but the numbers should be higher than 19.
  hence no such pairs exists..
• 9 years agoHelpfull: Yes(5) No(5)
• there are three possible co-prime pairs,but (1,60) is not possible because numbers are greater than 19.so ans will be (4,15) and (5,12)
• 10 years agoHelpfull: Yes(3) No(3)
• SAIRAM U ARE CRRECT RAUNAK U ARE WRONG;;;;;THE WRITE ANSWER IS 3
• 10 years agoHelpfull: Yes(2) No(1)
• 5 pairs like if u agree..
• 6 years agoHelpfull: Yes(0) No(0)