This MCQ module is based on: 6.4 Nature’s Favourite Sequence: The Virahānka–Fibonacci Numbers
TOPIC 22 OF 31
6.4 Nature’s Favourite Sequence: The Virahānka–Fibonacci Numbers
🎓 Class 7
Mathematics
CBSE
Theory
Ch 6 — Number Play
⏱ ~17 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 6.4 Nature’s Favourite Sequence: The Virahānka–Fibonacci Numbers
Targeting Class 7 level in Number Theory, with Basic difficulty.
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6.4 Nature's Favourite Sequence: The Virahānka–Fibonacci Numbers
Here is a sequence that appears again and again — in mathematics, in nature, in music, and in Sanskrit poetry:
\[1,\; 2,\; 3,\; 5,\; 8,\; 13,\; 21,\; 34,\; 55,\; 89,\; \ldots\]This is the Virahānka–Fibonacci sequence?. After the first two terms, each term is the sum of the previous two terms. For example, \(5 = 2 + 3\), \(8 = 3 + 5\), \(13 = 5 + 8\).
Discovery of the Virahānka Numbers
These numbers were first studied thousands of years ago by Indian scholars working on the rhythms of Sanskrit, Prakrit and other ancient Indian languages. Pingala, Bharata, Virahānka, Gopāla and Hemacandra all wrote about them — well before they became known in Europe.
Every syllable in a poem is either short (1 beat) or long (2 beats). The big question was: In how many ways can a rhythm have a given number of total beats?
Warm-up: An 8-beat rhythm can be written using short (1) and long (2) syllables — e.g., long long long long, or short short short short short short short short, or short long short long long short, and many more. How many in all?
A small example
Let \(n\) = total beats. We list the rhythms as addition-expressions using 1s and 2s.
| n (beats) | Different Ways | Number of Ways |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 1 + 1 | 2 | 2 |
| 3 | 1 + 1 + 1 | 1 + 2 | 2 + 1 | 3 |
| 4 | 1+1+1+1 | 1+1+2 | 1+2+1 | 2+1+1 | 2+2 | 5 |
| 5 | (see below) | 8 |
| 6 | — | 13 |
| 7 | — | 21 |
| 8 | — | 34 |
The 8 ways for \(n = 5\): 1+1+1+1+1, 1+1+1+2, 1+1+2+1, 1+2+1+1, 2+1+1+1, 1+2+2, 2+1+2, 2+2+1.
Definition — Virahānka–Fibonacci Numbers
The number of rhythms of \(n\) beats using 1s and 2s is the \(n^{th}\) Virahānka number \(V_n\). These satisfy \(V_n = V_{n-1} + V_{n-2}\), with \(V_1 = 1, V_2 = 2\).
Why does the recurrence hold?
Split the rhythms of \(n\) beats by their first syllable:
- If the rhythm starts with a short (1 beat), the remaining \(n - 1\) beats can be arranged in \(V_{n-1}\) ways.
- If it starts with a long (2 beats), the remaining \(n - 2\) beats can be arranged in \(V_{n-2}\) ways.
So \(V_n = V_{n-1} + V_{n-2}\). This is exactly the rule for the sequence!
Fig. 6.7 — Splitting rhythms by their first syllable gives the Virahānka recurrence.
Exploration — Write the next few terms
Write the next number in the sequence, after 55. Next: \(34 + 55 = 89\). Check with \(V_{n} = V_{n-1} + V_{n-2}\): \(V_{11} = 89\).
Write the next 3 numbers: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...
Write the next 3 numbers: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...
If you have to write one more number in the sequence above, could you tell it will be odd or even without adding the previous two?
Yes! Observe the parity pattern of the sequence: O, E, O, O, E, O, O, E, O, O, E, ... (one even after every two odd). Using this, the term after 233 will be odd + even = odd, and the next after that will be even + odd = odd, etc. So we can predict parity from the rule odd + even = odd, odd + odd = even.
Yes! Observe the parity pattern of the sequence: O, E, O, O, E, O, O, E, O, O, E, ... (one even after every two odd). Using this, the term after 233 will be odd + even = odd, and the next after that will be even + odd = odd, etc. So we can predict parity from the rule odd + even = odd, odd + odd = even.
What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities? The parity pattern OEOOEOOE... (period 3) repeats forever — because the recurrence "mixes" parities with fixed rules.
Virahānka Numbers in Nature
Count the petals of many common flowers. You will very often find a Virahānka–Fibonacci number:
- Daisies with 13 petals
- Daisies with 21 petals
- Daisies with 34 petals
The same numbers appear in pinecones, sunflower seeds, pineapples, ferns, nautilus shells, and even in the branching of rivers.
Fig. 6.8 — Petal counts of common flowers often match Virahānka–Fibonacci numbers.
History — Virahānka to Fibonacci
Pingala (around 200 BCE) first used the idea while studying Sanskrit metre. Virahānka (around 700 CE) stated the recurrence rule explicitly. Later, Gopāla (c. 1135 CE) and Hemacandra (c. 1150 CE) studied them in detail. Leonardo of Pisa (Fibonacci) introduced them to Europe about 1202 CE via his rabbit-pairs problem. That is why these are today properly called Virahānka–Fibonacci numbers.
Activity: Counting Rhythms of 6 beats
Activity: All 13 rhythms of 6 beats
L3 ApplyMaterials: Paper and pencil.
Predict: \(V_5 = 8\) and \(V_4 = 5\). So \(V_6\) should be \(8 + 5 = 13\). Can you list all 13 rhythms of 6 beats?
- Write the 8 rhythms of 5 beats, each prefixed with a short (1).
- Write the 5 rhythms of 4 beats, each prefixed with a long (2).
- Collect both groups: you should have \(8 + 5 = 13\) different rhythms.
Prefix with 1 (8 rhythms): 1+1+1+1+1+1, 1+1+1+1+2, 1+1+1+2+1, 1+1+2+1+1, 1+2+1+1+1, 1+1+2+2, 1+2+1+2, 1+2+2+1.
Prefix with 2 (5 rhythms): 2+1+1+1+1, 2+1+1+2, 2+1+2+1, 2+2+1+1, 2+2+2.
Total 13 = \(V_6\). ✓
6.5 Digits in Disguise — Cryptarithms
You have done many arithmetic operations with numbers. What if the digits were replaced by letters? Such puzzles are called cryptarithms? or alphametics — and they are great detective stories about parity, place-value and carries.
In the puzzles below, each letter stands for a particular digit (0 – 9). You have to figure out which digit each letter stands for.
Fig. 6.9 — A cryptarithm. Find the digits T and U.
Worked Example — T + T + T = UT
Here \(T + T + T = UT\) means a one-digit number added to itself three times gives a two-digit answer whose units digit is also \(T\).
So \(3T\) must end in the digit \(T\). Test T = 0, 1, 2, ..., 9:
- T = 0: \(3 \times 0 = 0\). Works but gives no 2-digit answer.
- T = 5: \(3 \times 5 = 15\) — units digit 5 = T ✓. So T = 5 and \(UT = 15\), giving U = 1.
So \(T = 5, U = 1\). Check: \(5 + 5 + 5 = 15\). ✓
What about U and T? Can T be 2? Can it be 3?
T = 2: \(3 \times 2 = 6\) — units digit 6 ≠ 2. Fail. T = 3: \(3 \times 3 = 9\) — units digit 9 ≠ 3. Fail. Only T = 5 makes \(3T\) a two-digit number ending in T. So T = 5, U = 1.
T = 2: \(3 \times 2 = 6\) — units digit 6 ≠ 2. Fail. T = 3: \(3 \times 3 = 9\) — units digit 9 ≠ 3. Fail. Only T = 5 makes \(3T\) a two-digit number ending in T. So T = 5, U = 1.
Another example — Two-digit sums
Consider: \(\overline{HM} + \overline{HM} = \overline{HMM}\). This is "HM added to itself equals HMM". Written as a 2-digit sum giving a 3-digit number.
Can you figure this one out?
\(HM + HM = 2 \times HM = HMM\). Let \(HM = 10H + M\). Then \(2(10H + M) = 100H + 10M + M\), so \(20H + 2M = 100H + 11M\), giving \(80H + 9M = 0\). Since H, M ≥ 0, the only solution is H = M = 0, which is trivial. So this particular version has no non-trivial solution — it is a good example of how parity and place-value can rule out most possibilities quickly.
\(HM + HM = 2 \times HM = HMM\). Let \(HM = 10H + M\). Then \(2(10H + M) = 100H + 10M + M\), so \(20H + 2M = 100H + 11M\), giving \(80H + 9M = 0\). Since H, M ≥ 0, the only solution is H = M = 0, which is trivial. So this particular version has no non-trivial solution — it is a good example of how parity and place-value can rule out most possibilities quickly.
More cryptarithm styles
Fig. 6.10 — Sample cryptarithms. Share with classmates and try different approaches.
Make a couple of cryptarithms yourself, exchange with a friend, and try each other's! Try to find puzzles with exactly one solution.
Competency-Based Questions
Scenario: Leena is decorating a rangoli border. She wants to arrange 8 beats of a traditional dance using short (1-beat) and long (2-beat) steps. She also wants to check if the number of petals in the flowers around her rangoli (3, 5, 8, 13, 21) follows the Virahānka rule.
Q1. How many different 8-beat dance sequences can Leena design? Justify.
L3 ApplyUse the recurrence: \(V_7 = 21, V_8 = 21 + 13 = 34\). So 34 different sequences are possible.
Q2. Leena notices that every third Virahānka number is even. Check: is this true for the first 12 terms? Predict whether \(V_{20}\) is even or odd.
L4 AnalyseParities of \(V_1, V_2, ..., V_{12}\): O, E, O, O, E, O, O, E, O, O, E, O. Evens occur at positions 2, 5, 8, 11 — every third term starting at n = 2. So \(V_{20}\): 20 = 2 + 6·3, i.e., position 20 = 2 + 18 → even. \(V_{20}\) is even.
Q3. Evaluate Leena's claim: "All numbers in the sequence of daisy petal counts (3, 5, 8, 13, 21) are Virahānka numbers, so any number of petals that is a multiple of 3 or 5 will also be a Virahānka number."
L5 EvaluateIncorrect. 6, 9, 10, 12, 15 are multiples of 3 or 5 but NOT Virahānka numbers. Being divisible by 3 or 5 is not the defining property; the recurrence \(V_n = V_{n-1} + V_{n-2}\) is.
Q4. Design a cryptarithm of the form \(A + A = BA\) where A and B are distinct digits. Solve your own puzzle and state why the solution is unique.
L6 CreateWe need \(2A\) to be a two-digit number whose units digit is A. That means \(2A - A = 10B\), i.e., \(A = 10B\). For B = 1, A = 10 — not a single digit. No solution. So instead try \(A + A = BA\) interpreted as \(2A\) has units digit A: so \(A \in \{0\}\) only (since \(2A \equiv A \pmod{10} \Rightarrow A \equiv 0\pmod{10}\)). So the cryptarithm has no non-trivial single-digit solution — the student can report that the puzzle is unsolvable, which is itself a valid answer. (Example of a solvable one: \(A + A = BB\) → A = 5, B = 1: \(5 + 5 = 11\)? No → \(A + A = BA\) style has no non-trivial solution; well-chosen constraints give uniqueness.)
Assertion–Reason Questions
Assertion (A): The 7th Virahānka number is 21.
Reason (R): Each term of the Virahānka sequence equals the sum of the previous two terms, starting with \(V_1 = 1, V_2 = 2\).
Reason (R): Each term of the Virahānka sequence equals the sum of the previous two terms, starting with \(V_1 = 1, V_2 = 2\).
Answer: (a). 1, 2, 3, 5, 8, 13, 21 — the 7th term is 21, computed by R.
Assertion (A): In the cryptarithm T + T + T = UT, the only non-zero solution is T = 5, U = 1.
Reason (R): 3T must end in the digit T, which happens only for T ∈ {0, 5} among single digits.
Reason (R): 3T must end in the digit T, which happens only for T ∈ {0, 5} among single digits.
Answer: (a). T = 0 gives 0 (no 2-digit result); T = 5 gives 15 = UT. R is the reason for A.
Assertion (A): The Fibonacci numbers were discovered entirely in Europe in the 13th century.
Reason (R): Leonardo of Pisa wrote about them in Liber Abaci (1202 CE).
Reason (R): Leonardo of Pisa wrote about them in Liber Abaci (1202 CE).
Answer: (d). A is false — Pingala, Virahānka, Gopāla and Hemacandra studied them in India centuries earlier. R is correctly describing Fibonacci's separate contribution.
Frequently Asked Questions — Number Play
What is Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool in NCERT Class 7 Mathematics?
Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 6: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool step by step?
To solve problems on Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 6: Number Play?
The essential formulas of Chapter 6 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool important for the Class 7 board exam?
Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool?
Common mistakes in Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool?
End-of-chapter NCERT exercises for Part 3 — Virahānka–Fibonacci Numbers & Digits in Disguise | Class 7 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.
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