What can the maximum number of digits be in the recurring block of digits in the decimal expansion of \( \frac{1}{17}\)? Perform the division to check your answer.

Asked by Vishal kumar | 2 years ago |  285

1 Answer

Solution :-

We need to find the number of digits in the recurring block of \( \frac{1}{17}\).

Let us perform the long division to get the recurring block of \( \frac{1}{17}\).

We need to divide 1 by 17, to get

We can observe that while dividing 1 by 17 we get 16 number of digits in the repeating block of decimal expansion which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that \( \frac{1}{17}\) = 0.0588235294117647..... or \( \frac{1}{17}\) = \( 0.\overline{0588235294117647}\), which is a non-terminating and recurring decimal.

Answered by Vishal kumar | 2 years ago

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