Question

When expanded as a decimal, the fraction $\frac{1}{7}$ has a repetend (the repeating part of the decimal) of $142857$. The last three digits of the repetend are $857$.



When expanded as a decimal, the fraction $\frac{1}{13}$ has a repetend that is $6$ digits long. If the last three digits of the repetend are $ABC$, compute the digits $A$, $B$, and $C$.

Answer 1

To find the decimal expansion of \(\frac{1}{13}\) and the last three digits of its repetend, we start by performing long division.

1. Calculate \(1 \div 13\):
- \(13\) goes into \(1\) \(0\) times. Thus, we write down \(0.\).
- We bring down \(1\), making \(10\). \(13\) goes into \(10\) \(0\) times. Write \(0\) again, so far we have \(0.0\).
- Bring down a \(0\), making \(100\). \(13\) goes into \(100\) \(7\) times (since \(13 \times 7 = 91\)). Write down \(7\).
- Subtract \(91\) from \(100\), getting \(9\).
- Bring down a \(0\), making \(90\). \(13\) goes into \(90\) \(6\) times (since \(13 \times 6 = 78\)). Write down \(6\).
- Subtract \(78\) from \(90\), getting \(12\).
- Bring down a \(0\), making \(120\). \(13\) goes into \(120\) \(9\) times (since \(13 \times 9 = 117\)). Write down \(9\).
- Subtract \(117\) from \(120\), getting \(3\).
- Bring down a \(0\), making \(30\). \(13\) goes into \(30\) \(2\) times (since \(13 \times 2 = 26\)). Write down \(2\).
- Subtract \(26\) from \(30\), getting \(4\).
- Bring down a \(0\), making \(40\). \(13\) goes into \(40\) \(3\) times (since \(13 \times 3 = 39\)). Write down \(3\).
- Subtract \(39\) from \(40\), getting \(1\).
- Bring down a \(0\), making \(10\). We observe we have returned to our original remainder, \(10\).

The decimal expansion we computed is:

\[
\frac{1}{13} = 0.\overline{076923}
\]

The repetend is \(076923\).

Next, we identify the last three digits of the repetend:

\[
ABC = 923
\]

We compute \( A \), \( B \), and \( C \):

- \( A = 9 \)
- \( B = 2 \)
- \( C = 3 \)

Finally, the answer is:

\[
\boxed{923}
\]