When expanded as a decimal, the fraction $\frac{1}{977}$ has a repetend (the repeating part of the decimal) that begins right after the decimal point, and is $976$ digits long. If the last three digits of the repetend are $ABC$, compute the digits $A$, $B$, and $C$.

To solve for the last three digits of the repetend of $\frac{1}{977}$, we need to determine the digits $A$, $B$, and $C$ from the repeating decimal expansion.

First, we recognize that the decimal expansion of $\frac{1}{977}$ is a repeating decimal with a period of 976 digits. This implies that:
\[
\frac{1}{977} = 0.\overline{d_1 d_2 \ldots d_{976}}
\]

We want to find the last three digits $d_{974}$, $d_{975}$, and $d_{976}$. Since the decimal is repeating, these digits are also the last three digits of the repetend.

To find these, we use the fact that:
\[
\frac{1}{977} = \frac{A}{10^n - 1} \quad \text{for some integer } A \text{ and } n = 976
\]

Given that:
\[
0.\overline{d_1 d_2 \ldots d_{976}} = \frac{1}{977}
\]

we can write the repetend as:
\[
d_1 d_2 \ldots d_{976}
\]

This periodicity implies:
\[
10^{976} \equiv 1 \pmod{977}
\]

Now, using modular arithmetic, we need to calculate the multiplicative inverse of $10$ modulo $977$, because the sequence of digits $d_1, d_2, \ldots$ is derived from powers of 10 reduced modulo 977.

We find $10^{975} \pmod{977}$ since we are interested in the last three digits:
\[
10^{975} \times 10 \equiv 1 \pmod{977} \implies 10^{975} \equiv 10^{-1} \pmod{977}
\]

Next, we need the modular inverse of $10$ modulo $977$. We solve for $x$ in:
\[
10x \equiv 1 \pmod{977}
\]

Using the Extended Euclidean Algorithm, we find:
\[
\begin{aligned}
977 &= 97 \times 10 + 7, \\
10 &= 1 \times 7 + 3, \\
7 &= 2 \times 3 + 1, \\
3 &= 3 \times 1 + 0.
\end{aligned}
\]

Working back up:
\[
\begin{aligned}
1 &= 7 - 2 \times 3, \\
3 &= 10 - 1 \times 7 \implies 1 = 7 - 2 \times (10 - 1 \times 7) = 3 \times 7 - 2 \times 10, \\
7 &= 977 - 97 \times 10 \implies 1 = 3 \times (977 - 97 \times 10) - 2 \times 10 \implies 1 = 3 \times 977 - 293 \times 10.
\end{aligned}
\]

Thus:
\[
-293 \times 10 \equiv 1 \pmod{977} \implies 10^{-1} \equiv 684 \pmod{977}.
\]

Hence:
\[
10^{975} \equiv 684 \pmod{977}
\]

Now we find the last three digits by considering:
\[
10^{974} \cdot 684 \pmod{977} \equiv 1 \pmod{977}
\]

Thus, the last three digits of the repetend of $\frac{1}{977}$ are $A = 6$, $B = 8$, and $C = 4$.

Therefore, $A = 6$, $B = 8$, and $C = 4$.

Discover how to compute the last three digits of the repetend (repeating part) of the decimal expansion for 1/977. This problem involves modular arithmetic, the extended Euclidean algorithm, and understanding repeating decimals. Suitable for advanced high school students and above.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation