Latest Elitmus Aptitude Question SOLUTION: 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?

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(#M40018624) ELITMUS Logical Reasoning question Reasoning Keep an EYE

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?

Asked In Elitmus Sagar Kadam (10 years ago)

Unsolved Read Solution (25)

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anjul meid  (10 years ago)

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

Is this solution Helpfull? Yes (79) | No (8)  

Ankit Goyal  (10 years ago)

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

Is this solution Helpfull? Yes (31) | No (1)  

Rajesh Kr Verma  (10 years ago)

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

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Hafiz Hussain  (10 years ago)

rajesh u missed 2070 so it is 31 in total

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pankaj kapoor  (10 years ago)

ans is 31 years

Is this solution Helpfull? Yes (10) | No (1)  

nishanth  (10 years ago)

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

Is this solution Helpfull? Yes (8) | No (19)  

vinod annam  (10 years ago)

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

Is this solution Helpfull? Yes (6) | No (16)  

Jitha Mitra Harivanam  (10 years ago)

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

Is this solution Helpfull? Yes (4) | No (7)  

siddhant  (10 years ago)

ans is 31.....

Is this solution Helpfull? Yes (4) | No   

Ashok   (9 years ago)

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

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SHIVSHANKAR KUMAR  (9 years ago)

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.

Is this solution Helpfull? Yes (4) | No   

RAHUL KUMAR  (9 years ago)

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

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febin zachariah  (10 years ago)

the answer is 32

Is this solution Helpfull? Yes (2) | No (7)  

govind bajaj  (9 years ago)

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

Is this solution Helpfull? Yes (2) | No   

sarab  (10 years ago)

i think 24..

Is this solution Helpfull? Yes (1) | No (6)  

rizwanur rahman  (10 years ago)

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)

Is this solution Helpfull? Yes (1) | No (6)  

SHIVSHANKAR KUMAR  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

Neha  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

apoorv agrawal  (10 years ago)

21 should bethe answer..

Is this solution Helpfull? Yes | No (8)  

apoorv agrawal  (10 years ago)

febin zachariah can u plez explain ur ans

Is this solution Helpfull? Yes | No (5)  

ankita  (10 years ago)

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

Is this solution Helpfull? Yes | No (11)  

Arun Kumar  (9 years ago)

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

Is this solution Helpfull? Yes | No (1)  

noracimons  (8 years ago)

I got this question in Elitmus yesterday. Answer is 32.

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Abhishek nkumar  (8 years ago)

I got this question in 30th august.and I marked option b 31..

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KISHAN KUMAR  (8 years ago)

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

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