The sum of four consecutive two-digit odd numbers , when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?
A. 21
B. 25
C. 41
D. 67
E. 73

• Perfect squares : 4,9,16,25,36,49,64,81......
  4=40/10,(40/4=10)(sum of 4 2-digit numbers cannot be 10).
  9=90/10,(90/4=22.5) (we cannot get decimas when we sum 4 2-digit nos)
  16=160/10,(160/4=40) (so the 2 digit numbers are around 40 , the 2- digit odd numbers around 40 are 37,39,41,43)
  
  
  37+39+41+43=160
  160/10=16 (a perfect square)
  
  Answer: 41
  
  
  When it comes to 36
  36=360/10, (360/4=90)
  
  The odd nums around 90 are 87,89,91,93
  
  87+89+91+93=360
  
  If we take 64 in the same way we get 640/4 = 160 (3-digit number so we cannot proceed more than this square number)
  
  Other than these all the other squares give decimal values when multiples by 10 and then divided by 4
• 9 years agoHelpfull: Yes(13) No(0)
• ans 41
  
  37+39+41+43=160/10=4*4
• 9 years agoHelpfull: Yes(6) No(1)
• its option C .41
  How to solve it:
  1. think of number which divided by 10 will be a whole number,
  2. Look at the option which is nearly equal to 1/4 th of that number.(because we are adding 4 consecutive numbers)
  
  here in this case:
  option 1: 21*4=80 (there is no perfect square when divided by 10 near by to 80)
  option 2: 25*4=100 (again no perfect square)
  option 3: 41*4=164 (there is an option 160 )
  160/4=16=4*4.
• 9 years agoHelpfull: Yes(4) No(0)
• sum of 4 terms in ap having cd=2
  s=2(2a+6),s/10=a perfect square(ps)
  =>2a+6=5*ps
  now putting different values of ps,a can be obtained
  ps=1,a=-1/2 not possible
  ps=4,a=7 not possible
  ps=9,a=39/2 not possible
  ps=16,a=37...so no may be 37,39,41,43,45
  and 41 is in the option too
• 9 years agoHelpfull: Yes(2) No(0)
• Lets number be a,a+2,a+4,a+6 where a is any 2 digit odd no
  so a+(a+2)+(a+4)+(a+6)=4a+12
  since
  4a+12 is divisible by 10 so its unit place digit must be 0(zero) or multiple of 10 i.e.(140,160,170,180)
  
  for 4a+12 becames multiple of 10 ,4a digit must contain 8 on unit place
  so only possible value of a is 17,27,37,47
  now u can check
  17*4+12=80 div 10 =8 not a perfect square
  27*4+12=120 div 10 =12 not a perfect square
  37*4+12=160 div 10 =16 it is a perfect square
  
  now odd number is a=37,a+2=39,a+4=41,a+6=43 so ans is 41(c) given in option
  
• 9 years agoHelpfull: Yes(2) No(0)
• A. 21
  
  21 + 22 + 23 + 24 = 90 , 90/10 = 9 [3*3=9]
• 9 years agoHelpfull: Yes(1) No(22)
• @ Roshan
  the no. should be consecutiv odd
  
• 9 years agoHelpfull: Yes(1) No(0)
• C. 41
  the last 4 one's digit must be 7,9,1,3 such that 0 comes at one's digit,
  so by doing it we get 37+39+41+43= 160/10= undroot16=4
• 9 years agoHelpfull: Yes(0) No(0)
• 41 is ans 37 39 41 43 = 160/10 =16 is prefect sqr
  
• 9 years agoHelpfull: Yes(0) No(0)
• WE ARE GETTING THE ANSWER via 21 also
  21+23+25+27/10=96/10=9
  so ans can be 21 also since,9 is also a perfect square
• 9 years agoHelpfull: Yes(0) No(2)