Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?

• 273*13 factor is=>3,7,13,13.
  (91,39) and (169,39).
  But if we check then only 39 and 91 satisfied the condition .
• 9 years agoHelpfull: Yes(10) No(0)
• Ans) 91
  HCF=13
  so by the given condition we get numbers as 65,78,91,104,117.130(which are multiples of 13(>60and
• 9 years agoHelpfull: Yes(4) No(1)
• the possible no between 60 and 140 are 65,78,91,104,117,130.
  if 273 is the lcm of both no than the no should be divisible by the no so there is only one option that is satisfying the condition .
  so ans is 91.
• 9 years agoHelpfull: Yes(2) No(0)
• HCF is 13 and LCM is 273, since one number is >60 and < 140, so multiples of 13 lies between 60 and 140 , but not multiples of 273.
  so, on enumber is definitely a multiple of 13. multiples of 13 between 60 and 140 are 65, 78, 91, 104, 117. and 91 is satisfying the HCF*LCM=one no. * other no. (formula)
  so answer is 91.
• 7 years agoHelpfull: Yes(1) No(0)
• HCF of two prime numbers is 1.
  Product of HCF and LCM = 1 x 319 = 319.product of 2 numbers = Product of their HCF and LCM
  p
  
  Now, co-primes with product 319 are (1,319) and (29,11)
  Since p > q, p = 29 and q = 11
  Then 3q - p = 33 - 29 = 4.
• 9 years agoHelpfull: Yes(0) No(1)
• Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?
• 7 years agoHelpfull: Yes(0) No(1)
• 15(5.8)+20(-3.1)=25 days
• 5 years agoHelpfull: Yes(0) No(0)