The number of natural numbers n such that (n+1)^2/(n+7) is an integer, is

i. 4
ii. 5
iii. 6
iv. None of these

• (n+1)^2/(n+7)
  =(n^2+2n+1)/(n+7)
  
  by simple algebraic polynomial division we get 36 as remainder
  now (n+1)^2/(n+7) to be an integer 36 have to divisable by (n+7)
  now 36 has 9 factors :1, 2, 3, 4, 6, 9, 12, 18 and 36
  and only 4 factors are greater then 7 : 9, 12, 18, 36
  so there are 4 natural numbers n such that (n+1)^2/(n+7) is an integer.
  n =(9-7) = 2 or,
  =(12-7)= 5 or,
  =(18-7)= 11 or,
  =(36-7)= 29
  like:
  when n=2
  (n+1)^2/(n+7)=3^2/9=1 is an integer
  when n=5
  (n+1)^2/(n+7)=6^2/12=3 is an integer
  when n=11
  (n+1)^2/(n+7)=12^2/18=8 is an integer
  when n=29
  (n+1)^2/(n+7)=30^2/36=25 is an integer
  
  so option (i) is correct
  there are 4 such numbers
• 9 years agoHelpfull: Yes(32) No(1)
• (n+1)^2/n+ 7 = k
  
  k= (n+7)^2 /n+7 – (12n + 48)/n+7
  
  if k is integer then (12 + 48)/n+7 has to be an integer.
  
  12n +48/n+7=12(n+4)/n+7.
  
  = 12 (n+7 -3)/n+7
  
  
  
  36/n+7 has to be an integer.
  
  n+7 = 9 ,12 , 18 , 36
  
  n can take 4 values
• 9 years agoHelpfull: Yes(8) No(0)
• Take the option and check one by one.
  n=4 then (4+1)^2/(7+4)=2.27 is not an integer.
  n=5 then (5+1)^2/(5+7)=3 is an integer.
  n=6 then (6+1)^2/(6+7)=3.76 is not an integer.
  So ans is ii. 5.
• 9 years agoHelpfull: Yes(4) No(9)
• we can use (n+1)^2 greater than (n+7) which gives n>2,-3........now there are four integers -2.-1,0,1 in between -3 and 2. so answer is 4.
• 9 years agoHelpfull: Yes(1) No(0)
• Use optional checking .. Take n=5
  Then (5+1)^2/(5+7)=36/12=3 I.e. an integer .
  So ans is 5.
• 9 years agoHelpfull: Yes(0) No(3)