Q2:

You know that $\frac{1}{7} = 0.\overline{142857}$. Can you predict what the decimal expansions of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ are, without actually doing the long division? If so, how? [Hint : Study the remainders while finding the value of $\frac{1}{7}$ carefully.]

Given Variables & Initial Setup

We are given the decimal expansion of the rational number $\frac{1}{7}$:

$\frac{1}{7} = 0.\overline{142857}$

The bar over the digits $142857$ indicates that this block of six digits repeats infinitely. We are tasked with predicting the decimal expansions of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7},$ and $\frac{6}{7}$ without performing long division, utilizing the properties of the remainders generated during the division of $1$ by $7$.

Step 1: Understanding the Cyclic Property of $\frac{1}{7}$

The number $142857$ is a well-known cyclic number. [Per number theory, a cyclic number is an integer in which cyclic permutations of the digits are produced when multiplied by successive integers]. Because $7$ is a prime number and the length of the repeating decimal block is $7 - 1 = 6$, the decimal expansions for any fraction $\frac{k}{7}$ (where $1 \le k \le 6$) will consist of the exact same six digits in the exact same cyclic order. The only difference is the starting digit.

Step 2: Analyzing the Remainders (The Hint)

When we perform the long division of $1 \div 7$, we append a zero to the remainder at each step to continue the division. The sequence of dividends and their corresponding quotients dictates the repeating block:

• $10 \div 7 = 1$ (Remainder $3$) → First digit is $1$
• $30 \div 7 = 4$ (Remainder $2$) → Second digit is $4$
• $20 \div 7 = 2$ (Remainder $6$) → Third digit is $2$
• $60 \div 7 = 8$ (Remainder $4$) → Fourth digit is $8$
• $40 \div 7 = 5$ (Remainder $5$) → Fifth digit is $5$
• $50 \div 7 = 7$ (Remainder $1$) → Sixth digit is $7$

[By the Division Algorithm], to find the decimal expansion of $\frac{k}{7}$, we simply look at the first step of dividing $k$ by $7$ (i.e., $10k \div 7$). The quotient gives us the starting digit of the cycle, and the rest of the digits follow the established cyclic order: $1 \rightarrow 4 \rightarrow 2 \rightarrow 8 \rightarrow 5 \rightarrow 7 \rightarrow 1$.

Step 3: Predicting the Decimal Expansions

Using the logic established in Step 2, we can predict the expansions through simple scalar multiplication or by identifying the starting digit:

Alternatively, this can be verified algebraically by multiplying the original repeating decimal by the respective numerators. For example:
$\frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714}$.
Because there are no carry-over digits that disrupt the cycle, the permutation holds perfectly true for all scalar multiples up to $6$.

Final Solution: Yes, the decimal expansions can be predicted without long division by utilizing the cyclic permutation of the digits $142857$. The predicted expansions are:
$\frac{2}{7} = 0.\overline{285714}$
$\frac{3}{7} = 0.\overline{428571}$
$\frac{4}{7} = 0.\overline{571428}$
$\frac{5}{7} = 0.\overline{714285}$
$\frac{6}{7} = 0.\overline{857142}$