The sum of four consecutive two-digit odd number,when devided by 10,becomes a perfect square.Which of the one can possibly be one of these four numbers?
a)21 b)25 c)41 d)67 e)73

• Let the four 2-digit odd numbers be
  n-3 n-1 n+1 n+3
  
  Sum of the 4 numbers ==> 4n
  
  acc to qn, when the sum is divided by 10
  we get a perfect square...
  perfect squares include==>1,4,9,16,25,36,49,.....
  
  Possible values of 4n/10 ==> 4, 16,36...
  
  If 4n/10=4
  n=10
  Hence, the corresponding nos are 7,9,11,13(all of which are NOT 2-digit nos)
  
  
  If 4n/10=16
  n=40
  Hence the corresponding nos are 37, 39, 41, 43
  
  If 4n/10=36
  n=90
  Hence the corresponding nos are 87, 89, 91, 93
  
  
  The answer to the ques therefore is Option C
• 12 years agoHelpfull: Yes(5) No(0)
• c)41
  sol: (37+39+41+43)/10=16, which is a perfect sqr.
• 12 years agoHelpfull: Yes(1) No(0)
• suppose two digit consecutive odd nos are is (10*y+x),(10*y+x+2),(10*y+x+4),
  (10*y+x+6)
  
  now sum of these no is=(40*y+4*x+12)and when diveded by 10 we have
  (40*y+4*x+12)/10
  
  by hit and trial we get 67 where x=7 and y=6(put it in equation
• 12 years agoHelpfull: Yes(0) No(0)
• Numbers can be
  
  21
  22
  23
  24
  ans 90
  divided by 10
  so answer is a
  
• 9 years agoHelpfull: Yes(0) No(0)