Answers

2015-12-22T09:49:11+05:30
Interval .......................||| No. of 1s

1-10 2
11-20 10
Now its clear that
21-30 , 31-40 , 41-50 , 51-60 , 61,70 , 71-80 , 81,90 , 91-99

So together all two digits will give = ( 2 + 10 + 1 * 8 ) = 20

Imagine the above groups as groups :

Group A : 1-10 , Group B: 11-20 , Group C : 21 - 99
We will use above information to find the rest
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3 Digit

Now 100 onwards to 999. ( Note from 100 , 200 , 300 , 400 , 500 , 600 , 700 , 800 , 900 = 9 terms )

Note : In each gap of 100 numbers ( ie 100-200 , 200-300 etc . There are : 1 ( 1-10 ) , 1 ( 11-20 ) , 1 ( 21 - 99 )

So

100 , 200 , 300 ..... 999 has 9 * Group A
has 9 * Group B
has 9 * Group C
Special cases : since 100 to 199 there is an extra 1 in hundreds place for each number so extra 1s = no.of 3digit numbers from 100 to 199 = 100

So total 1s from 3 digit numbers = 9*2 + 9*10 + 9*20 + 100
= 18 + 90 + 180 + 100
= 108 + 280 = 388
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4 digit : Thers 1 one in 1000

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Total = 20 + 388 + 1 = 409 .








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