Q. g(n)= 2 *sum of first n digit no+41 then wîch is the smallest non prime no. A)n=6 b)n=7 c)n=40 d) g(n) is always prime no.

• i think answer should be c)40 put n=40 in eqnt n(n+1)+41= 40*41+41= 1640+41=1681 which is not a prime no...because it square of 41...41*41=1681..and prime no are those no which is divisible by 1 or itself...g(n) is non prime at n=40
• 10 years agoHelpfull: Yes(19) No(2)
• d) g(n) is always prime no
  
  for any value n=1,2,3,... ,value of g(n) is always greater than 41
  so option a,b,c are incorrect
  
  g(n)= 2 *sum of first n digit no + 41
  g(1)=2*1+41=43
  g(2)=2*(10+11)+41 =83
  g(3)=3*(100+101+102)+41=647
  so g(n) is always prime no.
  
  
  
  however g(n) is not clear?
  
  
• 10 years agoHelpfull: Yes(9) No(8)
• ans is d
  g(n)=2*n(n+1)/2+41
  put the value n=1,2,3,4,4,10...
  2*1*2/2+41=43 its prime no
  2*10*11/2+41=151 its prime no
  
• 10 years agoHelpfull: Yes(0) No(4)
• a)n=6
  
  (n^2+2n+86)/2==>if n=6,then 134/2=67 which is prime no.
• 10 years agoHelpfull: Yes(0) No(1)
• n=40.... g(n)=1681 it is not prime
• 10 years agoHelpfull: Yes(0) No(0)
• 40 will be the ans
  g(n)=40*41+41
  then ans should be divided by 41
  so 40 will be the ans guys :)
• 10 years agoHelpfull: Yes(0) No(1)
• given g(n)=n*(n+1)+41 is prime number and n should small from given options n=6 6*7+42=83 so ans is n=6
• 10 years agoHelpfull: Yes(0) No(0)