Quant--Term

What is the 27393^rd term in the series 1234567891011121314……..?

a)4 b)3 c)5 d)6

Answer : c) 5

Solution :

By closely inspecting the series, one can find the following :
i) There are 9 single digit numbers starting from 1 to 9.
ii) After single digit numbers, 2 digit numbers are in sequence. There are 90 such numbers from 10 to 99. Number of digits occupied by all these numbers = 90 x 2 = 180 digits.
iii) Following 2 digit numbers there would be 3 digit numbers starting from 100 and proceeding upward till 999. There are 900 such numbers ranging from 100 to 999. Number of digits occupied by all these numbers = 900 x 3 = 2700

From above three findings from 1 to 999 the total number of digits occupied will be 9 + 180 + 2700 = 2889 . Following 999, there would be 4 digit numbers starting from 1000 and proceeding upwards.
To solve the question, we have to find 27393rd term from 1. In other words we have to find 27393 - 2889 = 24504^th term from 999.
Dividing 24504 by 4 we get 6126 leaving no remainder. This implies that the last digit of the 6126^th four digit number starting after 999 is our answer.
i.e last digit of 6126 + 999 = 7125 is our answer. Therefore our answer is 5.