what is the reminder when a 4 digit no p^2+17;where p is a prime no of form 6x+-1,

• considering +ve sign
  p^2+17 = (6x+1)^2 + 17 = 36x^2+1+12x +17 = 36x^2 + 12x + 18
  
  (p^2+17)/12 => (36x^2 + 12x + 18)/12 => rem = 6
  
  considering -ve sign
  p^2+17 = (6x-1)^2 + 17 = 36x^2+1-12x +17 = 36x^2 - 12x + 18
  
  (p^2+17)/12 => (36x^2 - 12x + 18)/12 => rem = 6
  
  for both cases rem = 6
  
  a) only 6
• 10 years agoHelpfull: Yes(46) No(0)
• Question was like this:
  
  P is a 4 digit prime number;when P^2 + 17 is divided by 12 what is the remainder?
  [Hint:- Prime number is denoted by 6x +- 1]
  
  Options:
  a) only 6
  b) only 4
  c) only 4&6
  d) cant be determined
• 10 years agoHelpfull: Yes(22) No(0)
• atleast give complete questions
  
• 10 years agoHelpfull: Yes(4) No(0)
• question is wrng.p is four digit prime number ,divided by 12.then what will be the remainder? ans will be 6
• 10 years agoHelpfull: Yes(1) No(1)
• IF u will take +ve sign then (6x+1)^2+17/12
  lets take x=2
  you will get rem=6
  same for -ve sign if you will do the same process but remember square value should be a prime no. you will get reminder 6.
  so optin.(a
  
• 8 years agoHelpfull: Yes(1) No(0)
• @rakesh pls expln how the rem 6 came...how did u simply
  p^2+17=(6x-1)^2+17=36x^2+1-12x+17=36x^2-12x+18
  
• 10 years agoHelpfull: Yes(0) No(0)
• mr. rakesh please explain with example
• 10 years agoHelpfull: Yes(0) No(1)