Question

How many digits are in the repeating cycle of the fraction 65/99?

Answer 1

To determine the number of digits in the repeating decimal cycle of the fraction \( \frac{65}{99} \), we first express it in decimal form.

We'll perform long division of 65 by 99:

1. Here, \( 65 < 99 \), so we note that the integer part is \( 0 \), and we need to consider the decimal expansion.
2. We rewrite \( 65 \) as \( 65.000000 \) to carry out the division.

When dividing:

- \( 99 \) into \( 650 \) goes \( 6 \) times because \( 99 \times 6 = 594 \).
- Subtract \( 594 \) from \( 650 \) to give \( 56 \).
- Bring down a \( 0 \) to make it \( 560 \).
- \( 99 \) into \( 560 \) goes \( 5 \) times because \( 99 \times 5 = 495 \).
- Subtract \( 495 \) from \( 560 \) to give \( 65 \).
- Bring down a \( 0 \) again to make it \( 650 \).

The process repeats from here since we are back at \( 650 \).

Thus, we have
\[
\frac{65}{99} = 0.65656565\ldots
\]
The repeating part is \( 65 \).

Therefore, to find the number of digits in the repeating decimal part, we count the digits in \( 65 \):
- There are **2 digits** in total.

In conclusion, the number of digits in the repeating cycle of the fraction \( \frac{65}{99} \) is
\[
\boxed{2}.
\]