29.the number 2^32 -1 has several divisors greater than 1 and less than 100.the sum of these divisors is clue:2^2n+1 is a prime for n=1,2,3,4 a)172 b)125 c)91 d)176

• To solve this qn we have use this fact-- (2^2n)+1 = prime for n=1,2,3,4 & the formula a2-b2 = (a+b)(a-b)
  Now
  (2^32 )-1 = (2^16 -1) (2^16+1) = (2^8 -1) (2^8+1)(2^16+1) = (2^4 -1) (2^4+1) (2^8+1)(2^16+1)
  Repeting the formula again & again we get:
  (2^32 )-1 = (2 -1) (2 +1)(2^2 +1) (2^4+1) (2^8+1)(2^16+1)
  = 1 x 3 x 5 x 17 x (factor larger than 100) x (factor larger than 100)
  So factors smaller than 100 are 3,5,17,15,51,85
  So sum = 3+5+17+15+51+85 = 176
• 8 years agoHelpfull: Yes(23) No(0)
• Consider the case of 2^5 = 32 .. 2x2x2x2x2x2 = 32 .... divisor of 32 will be 2, 4, 8 , 16 , 32 .. similarly for 2^6 = 64 divisor will be 2,4,8,16,32,64 ... now for 2^31 divisors between 1 and 100 will be 2,4,8,16,32,64 der sum will be 126 which is most nearby to answer 125.... so answer is 125
• 8 years agoHelpfull: Yes(3) No(3)
• Aditya ...It is not asking 2^31...It is asking (2^32)-1
• 8 years agoHelpfull: Yes(1) No(0)
• Hey AVIJIT can you please elobrate how you came up with 15,51 and 85?
• 8 years agoHelpfull: Yes(1) No(0)
• Thanx avijit
  
• 8 years agoHelpfull: Yes(0) No(0)
• bit confusing .... not getting the sol
• 8 years agoHelpfull: Yes(0) No(0)