20^2004 + 16^2004 - 3^2004 - 1 is divisible by????

• When dividend is of the form a^n+b^n or a^n-b^n:
  we have 3 rules
  Thrm 1:a^n+b^n is divisible by a+b when n is odd
  Thrm2: a^n-b^n is divisible by a+b when n is even
  Thrm3:a^n-b^n is always divisible by a-b
  Solution: 20^2004 + 16^2004-3^2004-1=(20^2004-3^2004)+(16^2004-1^2004)
  Now 20^2004-3^2004 is divisible by 17(Theorem 3)and 16^2004-1^2004 is divisible by 17 (Theorem 2)
  Hence the complete expression is divisible by 17
  20^2004+16^2004-3^2004-1=(20^2004-1^2004)+(16^2004-3^2004)
  Now 20^2004-1^2004 is divisible by 19 (Theorem 3) and 16^2004-3^2004 is divisible by 19 (Theorem 2).
  Hence the complete expression is also divisible by 19
  Hence the complete expression is divisible by 17 × 19 = 323
  
• 10 years agoHelpfull: Yes(34) No(5)
• divisible by 2..
  
• 10 years agoHelpfull: Yes(4) No(1)
• Ans:- 17
  
  (17+3)^2004 + (17-1)^2004 - 3^2004 - 1
  Use Binomial
• 10 years agoHelpfull: Yes(1) No(2)
• 20^2004=even no.
  16^2004=even no
  3^2004=odd no
  3^2004+1=even no.
  thus 20^2004+16^2004-(3^2004 +1)=even no.
  and every even no is divisible by 2,thus whole expression is divisible by 2.
• 10 years agoHelpfull: Yes(1) No(1)
• in 20^2004 unit digit is o
  in 16^2004 unit digit is 6
  in 3^2004 unit digit is 1
  so the unit digit in the expression in (0+6-1-1)=4
  so it is divisible by 2 or 4.
• 10 years agoHelpfull: Yes(1) No(1)
• please post options for these type of questions
• 10 years agoHelpfull: Yes(1) No(0)
• divisible by 4....
  
• 10 years agoHelpfull: Yes(0) No(2)
• 20^2004 + 16^2004 - 3^2004 - 1^2004
  then (20+16-3-1)^2004 so 32^2004
  Thus it is divisible by 2 and its multiples.
• 10 years agoHelpfull: Yes(0) No(2)