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

• ans 323
  (20^2004-3^2004)+(16^2004-1^2004) is divisible by (20-3)=17 and (16+1)=17
  (20^2004-1^2004)+(16^2004-3^2004)is divisible by (20-1)=19 and (16+3)=19
  hence the expression is divisible by=19*17=323
• 11 years agoHelpfull: Yes(13) No(2)
• @namrata mukherjee:
  i didnot get it...why did u do (16+1)and (16+3)?
  should it not be (16-1) and (16-3)..
  please clarify
• 11 years agoHelpfull: Yes(5) No(0)
• unit digit of no =0+6-1-1=4
  so is divisible by 2
  
• 11 years agoHelpfull: Yes(4) No(4)
• is there any logic ??? @DIPU
  
• 11 years agoHelpfull: Yes(1) No(0)
• ans 1
  (20^2004-1^2004)+(16^2004-3^2004)
  now a^n-b^n is always divisible by a-b weather n is even or odd so
  (20^2004-1^2004) and (16^2004-3^2004) is divisible by 19 and 13 and these 19 & 13
  are prime no. divisible by 1.
• 11 years agoHelpfull: Yes(1) No(3)
• 20^n+16^n-3^n-1^n
  (20+16-3-1)^n
  32^n
  ans will be 2,16,31,8,4
• 11 years agoHelpfull: Yes(1) No(3)
• here their is no options .....it is divisible by 2 and 323 ....by 2 it is power method .... gud sol namratha
  
• 11 years agoHelpfull: Yes(0) No(4)
• ans is (20-3)=17 and (16+1)=17,(20-1)and (16+3)=19
• 11 years agoHelpfull: Yes(0) No(0)