Class 9 Math Exercise 1.3 | Question 5

Question 5: What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \\frac{1}{17})? Perform the division to check.

Explanatory Answer | Exercise 1.3 Question 5

Property: The maximum number of digits in each repeating block of a recurring decimal will be one less than the divisor

Why?

The number of different remainders possible when a divisor x divides a sequence of numbers ranges from 0 to (x - 1). Excluding 0, we will have (x - 1) remainders.
Not convinced? List down all the different remainders you will get when integers from 10 to 20 are divided by 6. (You will learn more about it in Class 10 - Euclids Division Algorithm)
There will be a value of quotient corresponding to each of these remainders. The decimal places in the expansion of a recurring decimal are the quotients of the repeated divisions.
This means that the maximum possible number of decimal places in the repeating block of digits in the decimal expansion of a fraction is one less than the divisor.
Therefore, the maximum number of decimal places in the decimal expansion of \\frac{1}{17}) is 16.

Verifying by computing the decimal expansion of \\frac{1}{17})

\\frac{1}{17}) = 0.05882352941176470588235294117647...
You will notice that it actually has 16 digits in each repeating block of digits in the decimal expansion.
Note: The maximum number does not necessarily mean that you will always have (x - 1) digits. It could be lesser than that. For instance, \\frac{1}{6}) does not have 5 digits in each repeating block.