The digital sum of no is given as 1994 ie [1+9+9+4 = 23 then 2+3 =5]
unless we get a unit digit,
Find the number of perfect squares obtained in the digital sum from 2000 to 2199

optns r
22 42 16 nd may be d optns may be wrong but d ques is same

• Ans is 66
  There are only 3 single digit perfect square i.e. 1,4,9
  Total no. =200(including 2000 and 2199)
  
  when we sum 2000=2+0+0+0=2
  2001=2+0+0+1=3
  2002=2+0+0+2=4 and upto
  2199=2+1+9+9=21=2+1=3
  
  we see a pattern no. are going from 2,3,4---9,1,2and so on
  
  So no. of times pattern followed= 200/9=22 and 2 as remainder
  So xcept 2 and 3 each no. will come 22 times
  So total no. of times=22+22+22=66
  
• 8 years agoHelpfull: Yes(6) No(2)
• given 2000 and 2199 then if we start adding the digits from start i.e 2000 and proceed a sequence is produced i.e difference of 5 and 4
  
  here it goes
  
  2002= 2+0+0+2=4(perfect square)
  now
  
  (2002) + 5 = 2007
  taking 2007= 2+0+0+7=9(perfect square)
  
  now add +4 to 2007= 2011 which is 2+0+1+1=4(perfect square again)
  
  proceed this way...
• 8 years agoHelpfull: Yes(2) No(2)
• we have to find that how many times the number which are perfect square by following digital sum rule occurs in the series 2000 to 2199
  
  Total no. =200(including 2000 and 2199)
  
  when we sum 2000=2+0+0+0=2
  2001=2+0+0+1=3
  2002=2+0+0+2=4 and upto
  2199=2+1+9+9=21=2+1=3
  
  we see a pattern no. are going from 2,3,4,5,6,7,8,9,1,2and so on
  There are only 2 single digit perfect square i.e. 4,9
  200/9=22 reminder as 2
  so 22+22=44
• 8 years agoHelpfull: Yes(2) No(1)
• 44 is the correct answer
  
• 8 years agoHelpfull: Yes(1) No(3)
• but 31 and 26 are given option there.........
  what should be the?........
• 8 years agoHelpfull: Yes(1) No(3)
• 66.....after doing addition using digital method i.e 2+0+0+0=2
  2+0+0+1=3
  .......
  .....
  2+1+9+9=3
  ....so the number starts from 2,3,4...9,1,2,......9.....1,2,3
  for perfect square only 1,4,9 are there...so 2+64=66
• 8 years agoHelpfull: Yes(1) No(0)
• d.66 s the total no
• 8 years agoHelpfull: Yes(0) No(3)
• GuySs how to determine dat
  any formula involved
• 8 years agoHelpfull: Yes(0) No(1)
• as digital sum is single digit there are two possibilities either 4 or 9 i.e the sum of digits turn out to be 4 or 9.
  now, the maximum sum of digits can be 2+1+9+9= 21.
  
  So , we have to count all numbers whose sum of digits can be 4 or 9 (under the constraint that sum
• 8 years agoHelpfull: Yes(0) No(3)