5. We have seen that the repeating block of is a cyclic number. Try to find more numbers () whose reciprocals () produce decimals with repeating blocks that are cyclic.

Understand the Question A number's reciprocal produces a cyclic number when the decimal expansion repeats in a periodic block of length equal to . These numbers must be prime numbers, specifically known as full reptend primes (or long primes). To find more such numbers, we test prime numbers and check if the period of the repeating decimal for is . Step 1 · Examine the Reciprocal of 7 Dividing by The repeating block is with a period length of , which is . Thus, is a prime number with a cyclic repeating block. Step 2 · Find Other Full Reptend Primes We search for other prime numbers whose reciprocal has a repeating block of maximum possible length : For : Length of repeating block For : Length of repeating block For : Length of repeating block For : Length of repeating block $ Answer > Numbers whose reciprocals produce cyclic repeating blocks include (along with ). Common Mistakes Assuming all primes work: Not all prime numbers yield a full period of . For example, has a period of length instead of , and has a period of length . Omitting leading zeros in the repeating block: For fractions like , the leading zero after the decimal point is part of the 16-digit repeating block.

The World of Numbers | Exercise 3.5

Question 5

*5. We have seen that the repeating block of $\dfrac{1}{7}$ is a cyclic number. Try to find more numbers ( $n$ ) whose reciprocals ( $\dfrac{1}{n}$ ) produce decimals with repeating blocks that are cyclic.

Solution

Understand the Question

A number's reciprocal $\dfrac{1}{n}$ produces a cyclic number when the decimal expansion repeats in a periodic block of length equal to $n - 1$ .
These numbers $n$ must be prime numbers, specifically known as full reptend primes (or long primes).
To find more such numbers, we test prime numbers and check if the period of the repeating decimal for $\dfrac{1}{n}$ is $n - 1$ .

Step 1 · Examine the Reciprocal of 7

Dividing $1$ by $7$ $1 \div 7 = 0.142857142857\dots = 0.\overline{142857}$

The repeating block is $\overline{142857}$ with a period length of $6$ , which is $7 - 1$ . Thus, $7$ is a prime number with a cyclic repeating block.

Step 2 · Find Other Full Reptend Primes

We search for other prime numbers $n$ whose reciprocal $\dfrac{1}{n}$ has a repeating block of maximum possible length $n - 1$ :

For $n = 17$ : $1 \div 17 = 0.05882352941176470588\dots = 0.\overline{0588235294117647}$ Length of repeating block $= 16 = 17 - 1$
For $n = 19$ : $1 \div 19 = 0.0526315789473684210526\dots = 0.\overline{058823529411764721...} = 0.\overline{052631578947368421}$ Length of repeating block $= 18 = 19 - 1$
For $n = 23$ : $1 \div 23 = 0.04347826086956521739130434\dots = 0.\overline{0434782608695652173913}$ Length of repeating block $= 22 = 23 - 1$
For $n = 29$ : $1 \div 29 = 0.03448275862068965517241379310344\dots = 0.\overline{0344827586206896551724137931}$ Length of repeating block $= 28 = 29 - 1$$

Answer

Numbers whose reciprocals produce cyclic repeating blocks include $17, 19, 23, 29$ (along with $7$ ).

Common Mistakes

Assuming all primes work: Not all prime numbers yield a full period of $p - 1$ . For example, $\dfrac{1}{13} = 0.\overline{076923}$ has a period of length $6$ instead of $12$ , and $\dfrac{1}{11} = 0.\overline{09}$ has a period of length $2$ .
Omitting leading zeros in the repeating block: For fractions like $\dfrac{1}{17} = 0.\overline{0588235294117647}$ , the leading zero after the decimal point is part of the 16-digit repeating block.

Question 5

More questions in Exercise 3.5