The expression 2^6n - 4^2n, where n is the natural number is always divisble by ?

• ans: 48
  
  2^6n - 4^2n
  = (64^n - 16^n)
  
  (x^n - y^n) is always divisible by (x-y) for any natural no. n
  
  so, (64^n - 16^n) is always divisble by (64-16)=48
• 10 years agoHelpfull: Yes(2) No(0)
• As (x-y)is a factor of (x^n - y^n) for any natural number
  So (x^n - y^n) is always divisible by (x-y) for any natural number
  
  Now The expression 2^6n - 4^2n = (2^6)^n-(2^4)^n where n is the natural number Hence the expression 2^6n - 4^2n is always divisible by 2^6 - 2^4 = 48
• 10 years agoHelpfull: Yes(0) No(0)
• always divisible by 48; because by assuming n=1.... we can proceed as
  2^6-4^2=48
  so if you put n=1,2,3,4...so on then you will get the numbers divisible by 48
• 10 years agoHelpfull: Yes(0) No(0)
• ans is 48 [x^n-y^n is always divisable by x-y]
  
  64^n-16^n =64-16=48
• 10 years agoHelpfull: Yes(0) No(0)