Subject: General Aptitude · Chapter: Number System · Exam: 2010 · Marks: · Difficulty:
The sum of the digits of a natural number ( $10^{n}-1$ ) is 4707, where $n$ is a natural number. The value of $n$ is
(a)477
(b)523
(c)532
(d)704
Answer
Answer (as printed): B
Explanation
$10^{n}$ has $(n+1)$ digits. Then, 9 will appear $n$ times in $\left(10^{n}-1\right)$. So, sum of digits in $\left(10^{n}-1\right)=9 n$. $$\therefore \quad 9 n=4707 \Rightarrow n=\frac{4707}{9}=523 .$$
Explanation as extracted from the printed page; notation may be imperfect.