How to find the 28383 rd term of 1234567891011121314151617 ..... ?

Fingers crossed but is the answer 2?

i have done whole lot of calculations....

In that series, if you observe then 9th term is 9 (9)

191st term is 9 (99)

2891st term is 9 (999)

38891st term is 9 too (9999)

But, we desire the 28383rd term.. for that we know that 38891st number is 9999 (calculated it) and find the distance between 38891st term & 28383rd term

so it is 38891-28383 = 10508, but this is merely the distance... to find which number lies there we gotta divide 10508 by 4 (i hope you get why to divide it by 4)

10508/4 = 2627, this means that the number is 2627 away from 9999, so that number is 9999-2627 = 7372

So, the 28383rd term is 2, 28382nd term is 7 and so on.....

Plzz.. do correct me if I am wrong anywhere :)

All the best for CAT

Actually, it's infinitely consecutive numbers that written in simultaneously. hence:

1 digit number(1-9): 9 numbers

2 digit number(10-99): 90 numbers

3 digit number(100-999): 900 numbers

4 digit number(1000-9999):9000 numbers

5 digit number(10000-99999):90000 numbers, & so on

for your problem, we have 5 digit numbers. so first, we subtract it with the number of 1-4 digits numbers.

28383-9999=18384

18384:5(because it's 5 digit)=3676 R 4

3676th number from 10000= 13675--->last digit:5

4 digits after 5 is 1,3,6,&7-->4th number=7

So, the 28383rd term of 1234567891011121314151617 is 7

i hope it can help you