Question 44

The sum of digits of a natural number $$(10^{n}-1)$$ is 4707, where n is a natural number. The value of n is

Solution

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When 1 is subtracted from $$10^n$$, it always results in a series of 9s.

Let the number of 9's be n. 9(n)=4707 or n=523

Hence there are 523, 9's. Thus n=523

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