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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
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
