Latest TCS Aptitude Question SOLUTION: a person starts writing all 4 digits numbers.how many times had he written the digit 2??

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a person starts writing all 4 digits numbers.how many times had he written the digit 2??

Asked In TCS nagashree (11 years ago)

Solved sagar pahwa Read Solution (13)

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Read Solution (13) :

ravi  (11 years ago)

1)when 2 is at unit place=9*10*10*1
2)when 2 is at tenth place=9*10*1*10
3)when 2 is at hundred place=9*1*10*10
4)when 2 is at thousand place=1*10*10*10
so total no. of 2s=900+900+900+1000=3700

Is this solution Helpfull? Yes (97) | No (14)  

sagar pahwa  (11 years ago)

Number of 2's at units place(1000 to 9999)=900
Number of 2's at tenths place(1000 to 9999)=900
Number of 2's at hundreds place(1000 to 9999)=900
Number of 2's at thousands place(1000 to 9999)=1000

therefore total number of 2's =(900+900+900+1000)=3700

Is this solution Helpfull? Yes (39) | No (10)  

sums  (11 years ago)

Total 4 digit numbers that we can form is (9*10*10*10)
Total 4 digit numbers that we can form with out 2 is (8*9*9*9)
so 4 digit nos with 2 we can form (9*10*10*10)-(8*9*9*9) = 3168

In 3700 i feel there is redundancy becz nos are getting repeated ..
correct me if i am wrong ..

Is this solution Helpfull? Yes (23) | No (11)  

Rats  (11 years ago)

(1)At unit's digit = 10*10*9 = 900 (10 times in every 100)
(2)At ten's digit = 10*10*9 = 900 (10 times in every 100)
(3)At hundred's digit = 100*9 = 900 (100 times in every 1000)
(4)At thousand's digit = 1000 (2000 to 2999)
So total 900+900+900+1000=3700 times.....ans

Is this solution Helpfull? Yes (13) | No (2)  

Debdutto Chakraborty  (11 years ago)

No of times a digit repeats from 0 to 99 = 300

Going by that logic we have 3100 repetitions from 0 to 999

Therefore

from 1000 to 1999 = 3100
from 2000 to 2999 = 3100+1000 (Since every one contains 2)
from 3000 to 3999 = 3100
from 4000 to 4999 = 3100
from 5000 to 5999 = 3100
from 6000 to 6999 = 3100
from 7000 to 7999 = 3100
from 8000 to 8999 = 3100
from 9000 to 9999 = 3100

Therefore it comes to (3100*10)+1000 = 32,000

Is this solution Helpfull? Yes (7) | No (50)  

abhishek gupta  (11 years ago)

correct ans is: 4961.
case 1: when no contain only one digit of 2.
10*10*10+9*10*10*3=3700 i.e total 2's are=3700*1=3700.
case 2: when no contain two digits of 2.
10*10*3+9*10*3=570 i.e total 2's are=570*2=1140.
case 3: when no contain three digits of 2.
10*3+9=39 i.e total 2's are=39*3=117.
case 4: when no contain four digits of 2.
1 i.e total 2's are=1*4=4
hence total no of 2's are = 3700+1140+117+4 = 4961.

Is this solution Helpfull? Yes (6) | No (7)  

Shreyas Suryavanshi  (11 years ago)

2's at units place(1000 to 9999)=900
2's at tenths place(1000 to 9999)=900
2's at hundreds place(1000 to 9999)=900
2's at thousands place(1000 to 9999)=1000
2's =(900+900+900+1000)=3700

Is this solution Helpfull? Yes (5) | No (3)  

Mukesh Pavithran  (9 years ago)

When 2 is in unit place- 9*10*10*1= 900 Times.....
When it is in tens place 9*10*1*10=900 Times.......
When it is in hundred's place 9*1*10*10=900 Times.......
When it is in 1000's place 1*10*10*10=1000 Times.......
adding all the possibilities will give you 900+900+900+1000= 3700

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

Indrajit Bhandari  (8 years ago)

Answer is 3700
from 10-19 --> 1
from 20-29 --> 11
from 30-39 --> 1
from 40-49 --> 1
from 50-59 --> 1
from 60-69 --> 1
from 70-79 --> 1
from 80-89 --> 1
from 90-99 --> 1

which makes 20 from 0 to 99

now please observer that similar pattern follows for
000 - 099 -->20
100 - 199 --> 20
300 - 399 --> 20
400 - 499 --> 20
500 - 599 --> 20
600 - 699 --> 20
700 - 799 --> 20
800 - 899 --> 20
900 - 999 --> 20

however for 200 to 299 we have that extra 2 as first digit for all these 100 numbers, so
200 - 299 --> 120

so
000 - 999 -->300

so for
1000 - 1999 -->300
3000 - 3999 -->300
4000 - 4999 -->300
5000 - 5999 -->300
6000 - 6999 -->300
7000 - 7999 -->300
8000 - 8999 -->300
9000 - 9999 -->300

again for
2000 - 2999 -->300 + 1000 as we have that extra 2 in the first digit for all these numbers, hence

2000 - 2999 -->1300

So total number of times digit 2 is used 1000 - 9999 is 3700

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

jeevadhasan  (8 years ago)

WHICH IS THE CORRECT ANSWER

Is this solution Helpfull? Yes | No (1)  

srini  (8 years ago)

4digit number=1111......9999 so 2 divide so 3700 is divided by 4 digit number

Is this solution Helpfull? Yes | No (1)  

sai siddardha  (8 years ago)

1)when 2 is at thosand place=1*10*10*10
2)when 2 is at hundred place=8*1*10*10(since we have already used 2 in thousand place we can pic one number from remaining 8 ,suppose if we take 9*1*10*10 it is a mistake.for example in case 1(ie when 2 is at thousand place there is a possibility of occourence of a 4 digit number 2257 which is also possible in case 2 when we consider as 9*1*10 *2 as we can take any number from 1 to 9 inthousand place which can also be a 2 )
3)when 2 is at tens place=7*10*1*10(similar to the above explination )
4)when 2 is in units place=6*10*10*1
so total no. of 2s=600+700+800+1000=3100

Is this solution Helpfull? Yes | No (1)  

Ankur sharma  (6 years ago)

1)when 2 is at unit place=9*10*10*1
2)when 2 is at tenth place=9*10*1*10
3)when 2 is at hundred place=9*1*10*10
4)when 2 is at thousand place=1*10*10*10
so total no. of 2s=900+900+900+1000=3700

Is this solution Helpfull? Yes | No (1)  

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