29. Star mark question: x = ((2^32) – 1) the sum of factors of x which lie in between 1 and 100 is Hint: (2^2^n + 1) is a prime number for n= 1, 2, 3, 4.
a. 314
b. 176
c. 143
d. 431

• b. 176
  
  ((2^32) – 1) = (2^16+1)*(2^16-1)
  
  = (2^16+1)*(2^8+1)*(2^8-1)
  
  = (2^16+1)*(2^8+1)*(2^4+1)*(2^2+1)*(2^2-1)
  
  all 5 factors are prime,
  factors lying between 1 and 100 are : 3, 5, 15(3*5), 17, 51(3*17), 85(5*17)
  
  sum = 3+5+15+17+51+85 = 176
  
• 9 years agoHelpfull: Yes(20) No(0)
• ans is 432..
• 9 years agoHelpfull: Yes(3) No(2)
• according to the problem,
  (2^32 - 1) = (2^16 + 1)*(2^8 + 1)*(2^4 + 1)*(2^2 + 1)*(2^2 - 1)
  here all 5 factors are prime,
  so the divisors which lies between 1 to 100 are
  3 , 5 , 15 , 17 , 51 , 85
  whose summation is,
  3 + 5 + 15 + 17 + 51 + 85 = 176
• 9 years agoHelpfull: Yes(2) No(1)
• hw??? can u xplain plz............ supriya
• 9 years agoHelpfull: Yes(0) No(0)
• canu u xplain pls
  
• 9 years agoHelpfull: Yes(0) No(0)
• plz explain
• 9 years agoHelpfull: Yes(0) No(0)
• according to the problem,
  (2^32 - 1) = (2^16 + 1)*(2^8 + 1)*(2^4 + 1)*(2^2 + 1)*(2^2 - 1)
  here all 5 factors are prime,
  so the divisors which lies between 1 to 100 are
  3 , 5 , 15 , 17 , 51 , 85
  whose summation is,
  3 + 5 + 15 + 17 + 51 + 85 = 176...
• 9 years agoHelpfull: Yes(0) No(0)