Question 14
A recurring decimal is a decimal representation of a number whose digits are periodic in nature, meaning that, after some places, the same sequence of digits repeats continuously. For example, $\frac{1234}{9999}=0.12341234\ldots=0.\overline{1234}$. To calculate the sum of all the digits of a recurring decimal, aside from summing all the digits directly, there is another method as follows:
(a) It is known that the recurring decimal of $\frac1{17}$ is $16$ digits long, which you can assume to be true. To find the sum of all the digits of the recurring decimal of $\frac1{17}$, suppose $\frac1{17}=\frac{A}{10^{16}-1}$, where positive integer $A$ is the repetend. Since the period is $16$, $10^{16}-1$ is divisible by $17$ but $10^k-1$ is not divisible by $17$ for $1\le k<16$. Notice that $17$ is prime; by Fermat’s Little Theorem, $10^{16}\equiv1\pmod{17}$. Thus $10^{16/2}=10^8\equiv\pm1\pmod{17}$. Write the subsequent proof to show that the answer is $72$. (5 marks)
(b) It is known that the recurring decimal of $\frac1{179}$ is $178$ digits long, which you can assume to be true. What is the sum of all the digits of the recurring decimal of $\frac1{179}$? Use a similar method to part (a). (15 marks)
(a) It is known that the recurring decimal of $\frac1{17}$ is $16$ digits long, which you can assume to be true. To find the sum of all the digits of the recurring decimal of $\frac1{17}$, suppose $\frac1{17}=\frac{A}{10^{16}-1}$, where positive integer $A$ is the repetend. Since the period is $16$, $10^{16}-1$ is divisible by $17$ but $10^k-1$ is not divisible by $17$ for $1\le k<16$. Notice that $17$ is prime; by Fermat’s Little Theorem, $10^{16}\equiv1\pmod{17}$. Thus $10^{16/2}=10^8\equiv\pm1\pmod{17}$. Write the subsequent proof to show that the answer is $72$. (5 marks)
(b) It is known that the recurring decimal of $\frac1{179}$ is $178$ digits long, which you can assume to be true. What is the sum of all the digits of the recurring decimal of $\frac1{179}$? Use a similar method to part (a). (15 marks)