The expression 2^6n -4^2n, where n is a natural number is always divisible by

• {2^(6n) -4^(2n)} = [(64^n) - (16^n)] = [(16*4)^n - (16^n)] =[(16^n)*(4^n -1)];
  
  Now,for any natural number (16^n) is divisible 16
  
  and
  
  (4^n -1) is divisible by (4-1),i.e, by 3..
  
  So, the whole expression is always divisible (16*3)=48;
  
  So,The given expression is always divisible by (48) for all natural number n..
• 12 years agoHelpfull: Yes(13) No(5)
• The expression 2^6n -4^2n, where n is a natural number is always divisible by 2.
• 12 years agoHelpfull: Yes(1) No(7)