Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.). As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$. We obtain the non-repeating part 123 and the repeating part 63. Let's discuss why the non-repeating part appears here, and show that the fraction $\frac{1}{221}$ does not have a non-repeating part. The matter is that in the decimal representation of the fraction $\frac{34}{275}$, the digit 3 appears every time the remainder is 100 when dividing by 275. We see (and this is the key point!), that the same remainder 100 is given by different numbers: 650 and 1750. Where, in turn, did these 650 and 1750 come from? The number 650 was obtained by appending a zero to the number $r_{1}=65$ (the remainder from dividing 340 by 275). That is, $10 r_{1}=650$. Similarly, $10 r_{2}=1750$, where $r_{2}=175$. The numbers 650 and 1750 give the same remainder when divided by 275 because their difference is divisible by 275: $1750-650=10\left(r_{2}-r_{1}\right): 275$. This is possible only because the numbers 10 and 275 are not coprime. Now it is clear why the fraction $\frac{1}{221}$ does not have a non-repeating part: if $r_{1}$ and $r_{2}$ are different remainders from dividing by 221, then the product $10\left(r_{2}-r_{1}\right)$ is not divisible by 221 (the number 221, unlike 275, is coprime with 10 - the base of the number system, so there is no non-repeating part).

Thus, the decimal representation of the fraction $\frac{1}{221}$ has the form $\frac{1}{221}=0,\left(a_{1} a_{2} \ldots a_{n}\right)$. Let's find $n$. Denote $A=a_{1} a_{2} \ldots a_{n}$. Then $\frac{1}{221}=10^{-n} \cdot A+10^{-2 n} \cdot A+\cdots$. By the formula for

![](https://cdn.mathpix.com/cropped/2024_05_06_216317514e8812a39ee1g-4.jpg?height=248&width=343&top_left_y=230&top_left_x=1599)

825

1750

1650

$-\boxed{1000}$ $\frac{825}{-1750}$ the sum of an infinite geometric series $\frac{1}{221}=\frac{A}{10^{n}-1}$. From this, $A=\frac{10^{n}-1}{221}$. Since $A$ is a natural number, we need to find (the smallest) natural $n$ for which the number $10^{n}$ gives a remainder of 1 when divided by 221.

Note that $221=13 \cdot 17$. In general, an integer $B$ (in our case $B=10^{n}$) gives a remainder of 1 when divided by 221 if and only if $B$ gives a remainder of 1 when divided by both 13 and 17. The necessity is obvious. Sufficiency: if $B=13 k_{1}+1$ and $B=17 k_{2}+1$, then $13 k_{1}=17 k_{2}$, which means that $k_{1}$ is divisible by 17, i.e., $k_{1}=17 m$. Therefore, $B=13 \cdot 17 m+1$, and when divided by 221, the remainder is indeed 1. Now let's find such $n$ that the number $10^{n}$ gives a remainder of 1 when divided by 13. Consider the sequence $b_{n}=10^{n}$. Replace its terms with remainders when divided by 13. We get this: $b_{1}=10, b_{2}=9, b_{3}=13, b_{4}=3, b_{5}=4, b_{6}=1, \ldots$ Each subsequent term is uniquely determined by the previous one. Therefore, $\left\{b_{n}\right\}$ is a periodic sequence, where every sixth term is 1. We will do the same for 17. There, 1 will be equal to every 16th term. Thus, the remainder 1 when divided by both 13 and 17 will be obtained at $n=\operatorname{LCM}(6,16)=48$.

Answer: 48.
7. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25$ (712) is 3.)