HCF of 3240,3600 and a third number is 36 and their LCM is (2^4)x(3^5)x(5^2)x(7^2).
The third number is?

a> 2^2 x 3^5 x 7^2
b> 2^2 x 5^3 x 7^2
c> 2^5 x 5^2 x 7^2
d> 2^3 x 3^5 x 7^2

• as we know that... prodct of nmbrs always equal to the prodct of hcf and lcm
  so,
  3240*3600*unknown nmbr= 36*LCM,
  unknown nmbr= 36*lcm/(3240*3600),
  --------
  -----
  --
  -
  and i hope u can calculate next step to make ur answer,
• 11 years agoHelpfull: Yes(1) No(12)
• HCF = 36 = (2^2) x (3^2)
  LCM = (2^4) x (3^5) x (5^2) x (7^2)
  3240 = (2^3) x (3^4) x (5^1)
  3600 = (2^4) x (5^2) x (3^2)
  x = (2^2) x (3^5) x (7^2) ( a )
  
  Explanation:
  
  1.Finding power of 2:
  
  HCF is the product of least powers of common prime factors. The given product contains (2 ^ 2),
  which(2 ^ 2) is not contained by any of the given two numbers.Therefore it should come from x.
  
  2.Finding powers of 3:
  
  LCM is the product of highest powers of all the factors. The given product contains (3 ^ 5),
  which(3 ^ 5) is not contained by any of the given two numbers.Therefore it should come from x.
  
  3.Finding powers of 5:
  
  x should not contain powers of 5 as HCF does not contain it.
  
  4.Finding powers of 7:
  
  Since LCM contains 7 ^ 2 , which is not contained by any of the given two numbers.
  So, x should contain it.
• 5 years agoHelpfull: Yes(0) No(0)