Question

Let us suppose that: $n=1234567891011...787980$
The integer $n$ is formed by writing the positive integers in a row, starting with $1$ and ending with $80$, as shown above. Counting from the left, what is the $90$th digit of $n$?

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. $5$

Answer 1

Answer: (E)

$5$

$n=\mathrm{123456789.....}$ so

First $9$ even digit will be $1-9$

In next $20$ digits even digit will be $1$

In next $20$ digits even digit will be $2$

In next $20$ digits even digit will be $3$In next $20$ digits even digit will be $4$

ie $89$ digits.

So ${\left(90\right)}^{th}$ digit will be $5\left(50\right)$