Q5:

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 your answer.

Theoretical Foundation: Maximum Digits in a Repeating Block

To determine the maximum number of digits in the repeating block of the decimal expansion of a rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime integers), we analyze the possible remainders generated during the long division process.

[Per the Division Algorithm], when dividing by a divisor $q$, the only possible non-zero remainders are the integers from $1$ to $q - 1$. If a remainder of $0$ is reached, the decimal terminates. If the decimal is non-terminating and repeating, the sequence of remainders must eventually repeat. Because there are exactly $q - 1$ possible non-zero remainders, the division process can generate at most $q - 1$ distinct remainders before a previous remainder reappears, forcing the sequence of quotient digits to cycle.

For the fraction $\frac{1}{17}$:

• The divisor is $q = 17$.
• The maximum possible number of distinct non-zero remainders is $17 - 1 = 16$.

Therefore, the maximum number of digits in the repeating block of $\frac{1}{17}$ is strictly bounded by 16.

Step-by-Step Long Division Execution

To verify this theoretical maximum, we perform the long division of $1 \div 17$. We append zeros to the dividend and track the quotient digits and remainders at each step [applying $a = bq + r$].

At Step 16, we obtain a remainder of $1$. This is the exact same value we started with (the numerator of $\frac{1}{17}$). Because the remainder has repeated, the sequence of quotient digits will now repeat infinitely in the exact same order.

Visualizing the Cyclic Nature of Remainders

The following diagram illustrates the complete cycle of the 16 distinct remainders generated during the division process. The cycle proves that the repeating block has reached its maximum theoretical length.

Final Decimal Expansion

Collecting the quotient digits from Step 1 through Step 16, we construct the repeating decimal block:

$ \frac{1}{17} = 0.\overline{0588235294117647} $

Counting the digits under the vinculum (overline), we find exactly 16 digits.

Final Solution: The maximum number of digits in the repeating block of the decimal expansion of $\frac{1}{17}$ is 16. Performing the long division confirms that the repeating block is $0588235294117647$, which contains exactly 16 digits.