The sum of the digits of a natural number (10ⁿ - 1) is 4707, where n is a natural number. The value of n is :

A. 477

B. 523

C. 532

D. 704

Answer: Option B

Solution(By Examveda Team)

10ⁿ has (n + 1) digits, Then, 9 will appear n times in (10ⁿ - 1).
So, sum of digits in (10ⁿ - 1) = 9n
∴ 9n = 4707
⇒ n = $$\frac{4707}{9}$$  = 523