How many digits are in the repeating cycle of 1727

Question

?(1 point)

Answer 1

To determine the number of digits in the repeating cycle of $\frac{1}{1727}$, we can use the properties of repeating decimals.

The length of the repeating cycle of $\frac{1}{n}$ can be found by determining the smallest integer $d$ such that ${10}^d\equiv 1,(\text{mod},n)$ where $n=1727$.

First, we need to find the prime factorization of $1727$:

• Since $1727$ is odd and does not end in 0 or 5, it is not divisible by 2 or 5.
• Testing small primes, we find that $1727$ can be factored as $7\times 13\times 19$.

Next, we calculate the least common multiple of the lengths of the cycles of each prime factor separately:

1. For the prime $7$:

   • The order of $10,(\text{mod},7)$ is 6, since ${10}^1\equiv 3$, ${10}^2\equiv 2$, ${10}^3\equiv 6$, ${10}^4\equiv 4$, ${10}^5\equiv 5$, and ${10}^6\equiv 1$.

2. For the prime $13$:

   • The order of $10,(\text{mod},13)$ is 6, since ${10}^1\equiv 10$, ${10}^2\equiv 9$, ${10}^3\equiv 12$, ${10}^4\equiv 3$, ${10}^5\equiv 4$, and ${10}^6\equiv 1$.

3. For the prime $19$:

   • The order of $10,(\text{mod},19)$ is 18, since it takes 18 powers to return to 1.

To find the repeating cycle for $1727$, we need to take the least common multiple (LCM) of the orders found:

• LCM(6, 6, 18) = 18.

Thus, the length of the repeating cycle of $\frac{1}{1727}$ is 18 digits.