can nyone plz tell how to solve ds question with explanation (2^32 + 1 ) is completely divisible by whole no.. which of the no will completely divisible by this no ??
A) ( 2^16 + 1)
B) ( 2^16 -1)
C) 7 * 2^33
D) (2^96 + 1)

• let us consider 2^32 as x,therefore x+1
  now see option D,it is 2^96+1=x^3+1^3
  x^3+1^3=(x+1)(x^2-2*x+1)
  therefore,it is completely divisible by (2^96+1)
• 9 years agoHelpfull: Yes(5) No(0)
• By concept, x^n+a^n is divisible by x+a if n is an odd integer.
  Here Option D, 2^96+1 can be written as (2^32)^3 + 1^3 or (2^32)^3 + 1. If 2^32 is taken as x and 1 as a then we can rewrite it as x^3+1 which is divisible by x+a i.e 2^32+1.
• 9 years agoHelpfull: Yes(2) No(0)
• we can check last digit to check divisibility
  
  2^32 last digit is 16 +1= 17
  
  2^96 also last digit is 16+1=17
  
  so answer is D
• 9 years agoHelpfull: Yes(0) No(0)
• Ans:2^96+1
  2^32^3 + 1 = ( 2^32+1)*(2^32^2-1*2^32+1^2)
  if x+1 divides 2^31+1 then it will divide 2^32^3+1= 2^96+1
• 9 years agoHelpfull: Yes(0) No(0)