The World of Numbers | Exercise 3.5

Question 5

*5. We have seen that the repeating block of 17\frac{1}{7} is a cyclic number. Try to find more numbers (nn) whose reciprocals (1n\frac{1}{n}) produce decimals with repeating blocks that are cyclic.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution

We are looking for numbers whose reciprocals have a repeating block of length one less than the number itself.

Step 1 — Understanding 1/71/7

Let's look at the reciprocal of 7. We divide 1 by 7. 1÷7=0.142857142857...1 \div 7 = 0.142857142857... The repeating block is 142857\overline{142857}. The length of this block is 6. This length is 717-1.

So, 7 is a number with this property.\boxed{\text{So, } \textbf{7} \text{ is a number with this property.}}

Step 2 — Finding more numbers

We are looking for prime numbers nn. Their reciprocal 1/n1/n must have a repeating block. The length of this block must be n1n-1. These primes are called full reptend primes. Let's check some other prime numbers.

For 17: We divide 1 by 17. 1÷17=0.05882352941176470588...1 \div 17 = 0.05882352941176470588... The repeating block is 0588235294117647\overline{0588235294117647}. The length of this block is 16. This is 17117-1.

So, 17 is another such number.\boxed{\text{So, } \textbf{17} \text{ is another such number.}}

For 19: We divide 1 by 19. 1÷19=0.0526315789473684210526...1 \div 19 = 0.0526315789473684210526... The repeating block is 052631578947368421\overline{052631578947368421}. The length of this block is 18. This is 19119-1.

So, 19 is another such number.\boxed{\text{So, } \textbf{19} \text{ is another such number.}}

For 23: We divide 1 by 23. 1÷23=0.04347826086956521739130434...1 \div 23 = 0.04347826086956521739130434... The repeating block is 0434782608695652173913\overline{0434782608695652173913}. The length of this block is 22. This is 23123-1.

So, 23 is another such number.\boxed{\text{So, } \textbf{23} \text{ is another such number.}}

For 29: We divide 1 by 29. 1÷29=0.03448275862068965517241379310344...1 \div 29 = 0.03448275862068965517241379310344... The repeating block is 0344827586206896551724137931\overline{0344827586206896551724137931}. The length of this block is 28. This is 29129-1.

So, 29 is another such number.\boxed{\text{So, } \textbf{29} \text{ is another such number.}}

Answer

Numbers whose reciprocals produce cyclic repeating blocks are prime numbers nn where the length of the repeating block is n1n-1. Examples of such numbers are: (i) 7 (ii) 17 (iii) 19 (iv) 23 (v) 29

More questions in Exercise 3.5

Q1

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 720\frac{7}{20}, 415\frac{4}{15} and 13250\frac{13}{250}. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

Q2

Perform the long division for 113\frac{1}{13}. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 213\frac{2}{13}? Now compute 313\frac{3}{13}, 413\frac{4}{13}, etc. What do you notice?

Q3

Classify the following numbers as rational or irrational:

(i) 81\sqrt{81}

(ii) 12\sqrt{12}

(iii) 0.333330.33333 \dots

(iv) 0.1234512345123450.123451234512345 \dots

(v) 1.010010001000011.01001000100001 \dots (Notice the pattern: Is it repeating a single block?)

(vi) 23.56018561223987479012023.560185612239874790120

Find the explicit fractions in case they are rational.

Q4

The number 0.90.\overline{9} (which means 0.999990.99999\dots) is a rational number. Using algebra (let x=0.9x = 0.\overline{9}, multiply by 10, and subtract), explain why 0.90.\overline{9} is exactly equal to 1.

Q5

*5. We have seen that the repeating block of 17\frac{1}{7} is a cyclic number. Try to find more numbers (nn) whose reciprocals (1n\frac{1}{n}) produce decimals with repeating blocks that are cyclic.

← Back to The World of Numbers