S(n) is sum of first n natural numbers, Z(n) = 2S(n)+41, lowest value of n such that Z(n) is NOT a prime.

A.7 B.6 c. 40 d.always a prime

• c is the right answer
  as the sum of first N natural numbers is
  S(n)=n/2(1+n)
  so z(n)=2(s(n))+41=2*n/2(1+n)+41=n(1+n)+41
  
  Now put all the given options one by one
  take n=7
  z(7)=7(7+1)+41=56+41=97 a prime number
  take n=6
  z(6)=6(6+1)+41=41+41=83 again a prime number
  now take n=40
  z(40)=40(40+1)+41=40*41+41=41(40+1)=41*41=1681 not a prime number
  
  
• 8 years agoHelpfull: Yes(39) No(0)
• c)
  z(n) = 2*(n(n+1)/2) +41 = n(n+1) +41
  Substitute values from options one by one
  When n=40
  z(n) = 41*41 which is a non prime
  
• 8 years agoHelpfull: Yes(9) No(1)
• d)
  z(n)= 2n(n+1)/2 +41
  z(n)= n(n+1) +41
  substitute the values of n=0,1,2,3..
  the result is always a prime.
• 8 years agoHelpfull: Yes(7) No(14)
• hi anyone got the result for the exam on 13th march on hyd?? still i didn't get result.
• 8 years agoHelpfull: Yes(0) No(3)
• yes because lowest natural no is 1 and 1 is nt prime
• 8 years agoHelpfull: Yes(0) No(1)
• ans-d .
  all the option gives values which are not prime
• 8 years agoHelpfull: Yes(0) No(2)
• how 41+41 becomes 83 ? @ akansha
• 8 years agoHelpfull: Yes(0) No(0)
• as nancy said its the process and the option c is results also prime number which is 1681
• 8 years agoHelpfull: Yes(0) No(0)
• Since, S(n)=n*(n+1)/2
  
  and therefore, Z(n)=2*n*(n+1)/2 +41
  Z(n)=n(n+1) +41
  the only case where n=40, acc. to the options is satisfied
  Z(n)=40(41)+41;
  =41{40+1};
  =41*41;
  =1681, Because 1681 has divisors rather than 1 and itself. 1681 can be divided by 41 and . Prime factorization of 1681 (list of divisors): 41
• 8 years agoHelpfull: Yes(0) No(1)
• d.always a prime
• 8 years agoHelpfull: Yes(0) No(2)