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
This Question Belongs to Arithmetic Ability >> Number System
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