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 $C$.

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 $C$.

GPT-4o mini · Answer

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: $$ \boxed{923} $$