Digital sum of 2007 is 9.
2+0+0+7 = 9 (which is square of 3)

In between 2000-2199, how many years are there which digital sum are square of any digit?

• answer is 31...first categorise by digital sum i.e 4,9,16 so we get 5 numbers having digital sum 4 ,15 numbers having digital sum 9 and 11 numbers having digital
  sum 16 adding we get 5+15+11=31 years
• 11 years agoHelpfull: Yes(79) No(8)
• 2002 1
  2007 2
  2011 3
  2016 4
  2020 5
  2025 6
  2034 7
  2043 8
  2052 9
  2059 10
  2061 11
  2068 12
  2070 13
  2077 14
  2086 15
  2095 16
  2101 17
  2106 18
  2110 19
  2115 20
  2124 21
  2133 22
  2142 23
  2149 24
  2151 25
  2158 26
  2160 27
  2167 28
  2176 29
  2185 30
  2194 31
  
  Ans = 31
• 11 years agoHelpfull: Yes(31) No(1)
• digital sum of the 2199=2+1+9+=21
  so 16 is the least square digit in the given range if the year.
  i.e we have 4(sq of 2),9(Sq of 3), 16(sq of 4)
  In 2000-2199. The 2 is constant means 2--- in between 2000 - 2199
  I.e 4=2+2
  9=2+7
  16=2+14
  we have to find the 2 digit number less than 100 whose sum is either 2,4,14
  in the categories
  4(2+2)=== 2002,2020,2011.
  9(2+7)===2007,2016,2061,2025,2052,2034,2043.
  16(2+14)===2059,2095,2068,2086,2077.
  Now same between 2100-2199.
  21 is const the 2+1=3..
  ie 4(3+1)===2101,2110.
  9(3+6)===2106,2160,2115,2151,2124,2142,2133.
  16(3+13)===2149,2194,2158,2185,2167,2176
  TOTAL YEAR=30
• 11 years agoHelpfull: Yes(26) No(30)
• rajesh u missed 2070 so it is 31 in total
• 11 years agoHelpfull: Yes(24) No(0)
• ans is 31 years
• 11 years agoHelpfull: Yes(10) No(1)
• 2 remains constant as the number starts from 2
  possible values of second digit=0,1
  
  so for getting total sum as 4= we calculate for 2 as our first digit is 2 itself
  to check possibilities carry 2 steps check for numbers with 20 first and then 21
  possibilities=2020,2002,2110,2011
  
  getting total sum as 9=we calculate for 7 as our first digit is 2
  possibilities=2007,2070,2052,2043,2034,2025,2061,2016,2160
  possibilities for 16=we calculate for 14
  =2077,2059,2095.2086,2068,2194,2149,2185,2158,2176,2167,
  
  ans:24
• 11 years agoHelpfull: Yes(8) No(19)
• for 4:
  2002,2020,2101,2110,2011
  (total 5)
  
  for 9:
  2007,2070,2043,2034,2052,2025,2016,2061,2160,2133,2142,2124,2151,2115
  (total 14)
  
  for 16:
  
  2059,2095,2077,2086,2068, 2137,2173,2167,2176,2158,2185,2149,2194
  (total 13)
  
  so total years is 5+14+13=32
  
  
• 11 years agoHelpfull: Yes(6) No(16)
• RAJESH is correct I think ans is 30
  4, 9 and 16 will be square possible in 2,7,14 terms
  so for 4: 20(0,2)-2 ways
  (1,1)-1
  21(0,1)-2
  so for 9: 20(0,7)-2
  (1,6)-2
  (2,5)-2
  (3,4)-2
  21(0,6)-2
  (1,5)-2
  (2,3)-2
  so for 16:20(9,5)-2
  (8,6)-2
  (7,7)-1
  21(9,4)-2
  (8,5)-2
  (7,6)-2
  therefore in total 30 years
• 11 years agoHelpfull: Yes(4) No(7)
• ans is 31.....
• 11 years agoHelpfull: Yes(4) No(0)
• sum of 2199 is 21.so under 21, 4,9 and 16 i perfect squire.but thousand digit is 1. so always take care of this.
  For 4:-take 0,0,2 makes two number
  0,1,1 makes three number
  total no are 5 for 4.
  for 9:-0,0,7 makes 2 no.
  0,1,6 4
  0,2,5 2
  0,3,4 2
  1,1,5 2
  1,2,4 2
  1,3,3 1
  total no are 15 for 9
  For 16 :-takes 0,7,7 makes no 1
  0,8,6 2
  0,9,5 2
  1,7,6 2
  1,8,5 2
  1,9,4 2
  total no are 11 for 16
  so,sum of 5,15,11 is 31
• 10 years agoHelpfull: Yes(4) No(0)
• digital sum of the 2199=2+1+9+=21
  so 16 is the least square digit in the given range if the year.
  i.e we have 4(sq of 2),9(Sq of 3), 16(sq of 4)
  In 2000-2199. The 2 is constant means 2--- in between 2000 - 2199
  I.e 4=2+2
  9=2+7
  16=2+14
  we have to find the 2 digit number less than 100 whose sum is either 2,4,14
  in the categories
  4(2+2)=== 2002,2020,2011,2101.
  9(2+7)===2007,2016,2061,2025,2052,2034,2043.
  16(2+14)===2059,2095,2068,2086,2077.
  Now same between 2100-2199.
  21 is const the 2+1=3..
  ie 4(3+1)===2101,2110.
  9(3+6)===2106,2160,2115,2151,2124,2142,2133.
  16(3+13)===2149,2194,2158,2185,2167,2176
  TOTAL YEAR=31.
• 10 years agoHelpfull: Yes(4) No(0)
• ans=31
  4 which is 2^2(2002,2020,2011,2110,2101)
  9 which is 3^2(2025,2052,2007,2070,2061,2016,2043,2034,2160,2106,2124,2142,2133,2151,2115)
  16 which is 4^2(2095,2059,2077,2086,2068,2185,2158,2176,2167,2194,2149)
  so total no of years=5+15+11=31
• 10 years agoHelpfull: Yes(3) No(0)
• the answer is 32
• 11 years agoHelpfull: Yes(2) No(7)
• the square of a no is 4, 9 16(maximum)
  so first the sum is 4 :- 2002,2011,2020,2101
  second some is 9 so year is: 2007,16,34,25,43,52,61,70,2106,2115,2133,2124,2133,2142,2151,2160,2133
  and sum is 16 is:= 2059,68,77,86,95,49,58,67,76,85,94
  so total no of year is +++======= 31 ans
• 10 years agoHelpfull: Yes(2) No(0)
• i think 24..
• 11 years agoHelpfull: Yes(1) No(6)
• the answer is 19. the limit to the year is 2199. so maximum sum of digits is 2+1+9+9= 21. hence the numbers whose squares are to be found are 2,3 and 4 as their squares are 4,9 and 16 which are less than 21. going by this procedures the valid years are 2101(2^2),2110(2^2),2020(2^2),2011(2^2),2002(2^2),2007(3^2),2016(3^2),2061(3^2),2160(3^2),2106(3^2),2025(3^2),2052(3^2),2034(3^2),2043(3^2),2059(4^2),2068(4^2),2077(4^2),2095(4^2),2086(4^2)
• 11 years agoHelpfull: Yes(1) No(6)
• digital sum of the 2199=2+1+9+=21
  so 16 is the least square digit in the given range if the year.
  i.e we have 4(sq of 2),9(Sq of 3), 16(sq of 4)
  In 2000-2199. The 2 is constant means 2--- in between 2000 - 2199
  I.e 4=2+2
  9=2+7
  16=2+14
  we have to find the 2 digit number less than 100 whose sum is either 2,4,14
  in the categories
  4(2+2)=== 2002,2020,2011,2101.
  9(2+7)===2007,2016,2061,2025,2052,2034,2043.
  16(2+14)===2059,2095,2068,2086,2077.
  Now same between 2100-2199.
  21 is const the 2+1=3..
  ie 4(3+1)===2101,2110.
  9(3+6)===2106,2160,2115,2151,2124,2142,2133.
  16(3+13)===2149,2194,2158,2185,2167,2176
  TOTAL YEAR=31
• 10 years agoHelpfull: Yes(1) No(0)
• for 4: 2002, 2020, 2101, 2110, 2011 (total 5)
  for 9: 2007, 2070, 2043, 2034, 2052, 2025, 2016, 2106, 2061, 2160, 2133, 2142, 2124, 2151, 2115
  (total 15)
  for 16: 2077, 2059, 2095, 2086, 2068, 2194, 2149, 2185, 2158, 2176, 2167
  (total 11)
  
  Total = 5+15+11 = 31
• 10 years agoHelpfull: Yes(1) No(0)
• 21 should bethe answer..
  
• 11 years agoHelpfull: Yes(0) No(8)
• febin zachariah can u plez explain ur ans
  
• 11 years agoHelpfull: Yes(0) No(5)
• as the last limit is 2199 ...the sum of the digits of 2199 is 21 thus the possible squares can be 4,9,16...now we have the thousands digit reserved by 2 ,thus for the sum to be 4 we can have 2002,2020,2011,2101,2110;for 9 we an have 2007,2070,2043,2034,2016,2061,2106,2160,2052,2025;for 16 we can have 2077,2086,2068,2059,2095;thus in total there are 20 such numbers possible
• 11 years agoHelpfull: Yes(0) No(11)
• Well
  a+b+c+d
  a=2 already
  b=0,1
  4,9,16 can only be few squares
  testing each case
  b=0
  3,8,5 sol of c,d respectively
  b=1
  2,7,6 sol of c,d respectively
  
  total 31
• 10 years agoHelpfull: Yes(0) No(1)
• I got this question in Elitmus yesterday. Answer is 32.
• 9 years agoHelpfull: Yes(0) No(0)
• I got this question in 30th august.and I marked option b 31..
• 9 years agoHelpfull: Yes(0) No(0)
• ANS--31
  
  
  1ST---FOR 4
  ===============================================================
  2002-2020-2011-2110-2101-->(TOTAL=5)
  
  2nd FOR 9
  =================================
  2007-2106-2016-2015-2034-2043-2052-2061-2070-2160-2124-2142-2133-2115-2151--->(TOTAL=15)
  
  3rd FOR 16
  ========================================================================
  2077-2059-2095-2086-2068-2194-2149-2185-2158-2176-2167---->(TOTAL=11)
  
  FINAL=5+15+11=31ANS
• 9 years agoHelpfull: Yes(0) No(0)