Latest TCS Aptitude Question SOLUTION: A person starts writing all the 4 digits number , how many times he has written the digit 2 ?

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A person starts writing all the 4 digits number , how many times he has written the digit 2 ?

Asked In TCS hima (9 years ago)

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RAKESH  (9 years ago)

---2 => 9*10*10 = 900 numbers contains 2 at unit place

--2- => 9*10*10 = 900
-2-- => 9*10*10 = 900
2--- => 10*10*10 = 1000

total = 900 + 900 + 900 + 1000 = 3700

Is this solution Helpfull? Yes (28) | No (5)  

krishna kanth  (9 years ago)

2---=>10*10*10=1000 numbers contains 2 in thousands place
-2-- =>here 2 and 0 wont come in thousands place
=>8*10*10=800
--2- => now 2 wont come in thousands place and hundreds place and 0 wont come in thousands place
=>8*9*10=720
---2 => 0 wont come in thousands place and 2 will not come in any of the remaining places
8*9*9=648
total=1000+800+720+648=3168

Is this solution Helpfull? Yes (6) | No (2)  

Ajay Singh Yadav  (9 years ago)

all 4 digit no =9*10*10*10=9000
4 digit no not contains 2=8*9*9*9=5832
4 digit no contains 2=9000-5832=3168

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Ajay Singh Yadav  (9 years ago)

sorry bt its how many times digit 2 so
-------------------
2 one time
2 at unit place=1*9*9*9=729
2 at any other place=8*9*9*1*3!/2!=1944
total 2 encountered=2673 times
----------------
2 two times
2 at unit place=1*9*9*1*3!/2!=243
2 at any other place=8*9*1*1*3!/2!=216
total 2 encountered=458*2=918 times
----------------
2 three times
2 at unit place=1*1*1*9*3!/2!=27
2 at any other place=8*1*1*1=8
total 2 encountered=35*3=105 times
----------------------
2 four times=1
total 2 encountered=4*1=4 times
-----------------
now total 2 encountered=2673+918+105+4=3700

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sks  (9 years ago)

If all no. starting from 1 to (10^n)-1 (i.e. up to n digits)
have to be written then we need to write n*10^(n-1) times
a non zero digit.
so for only 4 digits, required no. is (4*10^3)-(3*10^2)=3700.

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