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Themed maths · 23 November

Fibonacci Day maths for CBSE Class 10

23 November is Fibonacci Day: written month first, 11/23 reads 1, 1, 2, 3. This ready-to-teach lesson has a 5-minute starter, a 35-minute main activity of Fibonacci puzzles, Euclid’s division algorithm and the golden ratio, an extension and full worked answers. No calculator.

Level
CBSE Class 10 (Class 9 can try most of it)
Time
40 minutes, plus a 10-minute extension
Topics
Number patterns and sequences; Real numbers: Euclid’s division algorithm, HCF and LCM; Pair of linear equations; Quadratic equations and surds
Equipment
No calculator. Where a decimal is wanted, the question gives the value to use.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minFibonacci-type puzzles
Main: task B12 minEuclid’s division algorithm
Main: task C11 minThe golden ratio
Extension10 minFast finishers or homework

Starter (5 minutes)

Quick-fire mental maths. In the Fibonacci sequence 1, 1, 2, 3, 5, … each term is the sum of the two before it.

  1. Write down the next three terms: 1, 1, 2, 3, 5, 8, …
  2. Find the HCF of 21 and 34.
  3. Solve x2 − x − 1 = 0 using the quadratic formula.
  4. Find the sum of the first 7 Fibonacci numbers and compare it with the 9th.

Main activity (35 minutes)

Task A: Fibonacci-type puzzles (12 min)

In a Fibonacci-type sequence you choose the first two terms; after that, each term is the sum of the two before it.

  1. The first two terms are 3 and 7. Find the 8th term.
  2. A number trick: the sum of the first 10 terms of any Fibonacci-type sequence is 11 times the 7th term. Using first terms a and b, prove it. Then find the sum of the first 10 terms of the sequence in part (a) in one step.
  3. A Fibonacci-type sequence has 3rd term 12 and 6th term 50. Find the first two terms.

Task B: Euclid’s division algorithm (12 min)

Use a = bq + r, 0 ≤ r < b, again and again until the remainder is 0.

  1. Find HCF(144, 89) using Euclid’s division algorithm, and count the division steps. What do you notice about the remainders?
  2. Find HCF(144, 21). Both are Fibonacci numbers: 144 is the 12th and 21 is the 8th. What do you notice?
  3. Find the LCM of 21 and 34.

Task C: The golden ratio (11 min)

The golden ratio φ is the positive root of x2 = x + 1. Use √5 = 2.236.

  1. Find φ exactly, and then to 3 decimal places using √5 = 2.236.
  2. Show that φ2 = φ + 1 by working out both sides exactly.
  3. Divide 34 by 21, and 55 by 34, each to 3 decimal places. Compare with φ.

Extension (10 minutes)

Counting puzzles where Fibonacci numbers appear.

  1. You climb a staircase taking 1 or 2 steps at a time. There is 1 way to climb 1 step, 2 ways to climb 2 steps and 3 ways to climb 3 steps. How many ways are there to climb 10 steps?
  2. How many ways are there to tile a 2 by 8 strip with 2 by 1 dominoes?

For teachers

Teacher notes and full worked answers
  • Fibonacci Day is 23 November because 11/23, written month first, reads 1, 1, 2, 3.
  • The number trick in task A (b) is a good opener: ask a student for two numbers, write ten terms and give the total before they finish adding.
  • Task B (a) shows why consecutive Fibonacci numbers make Euclid’s algorithm as slow as possible: every quotient is 1.

Starter

  1. 13, 21, 34
    • 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34
  2. 1
    • 21 = 3 × 7 and 34 = 2 × 17 have no common factor except 1.
  3. x = (1 ± √5)/2
    • x = (1 ± √(1 + 4))/2 = (1 ± √5)/2
  4. 33, which is one less than 34
    • 1 + 1 + 2 + 3 + 5 + 8 + 13 = 33
    • The 9th term is 34.

Task A: Fibonacci-type puzzles

  1. 115
    • 3, 7, 10, 17, 27, 44, 71, 115
  2. 55a + 88b = 11(5a + 8b); 781
    • The terms are a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, 5a + 8b, 8a + 13b, 13a + 21b, 21a + 34b.
    • They add to 55a + 88b = 11(5a + 8b), and 5a + 8b is the 7th term.
    • For 3, 7, …: the 7th term is 71, so the sum is 11 × 71 = 781.
  3. 5 and 7
    • 3rd term a + b = 12; 6th term 3a + 5b = 50.
    • Multiply the first by 3: 3a + 3b = 36. Subtract: 2b = 14, so b = 7 and a = 5.

Task B: Euclid’s division algorithm

  1. HCF = 1, in 10 steps; the remainders are the Fibonacci numbers going down
    • 144 = 89 × 1 + 55, 89 = 55 × 1 + 34, 55 = 34 × 1 + 21, 34 = 21 × 1 + 13, 21 = 13 × 1 + 8,
    • 13 = 8 × 1 + 5, 8 = 5 × 1 + 3, 5 = 3 × 1 + 2, 3 = 2 × 1 + 1, 2 = 1 × 2 + 0
    • 10 steps, and the last non-zero remainder is 1.
  2. 3, which is the 4th Fibonacci number, and HCF(12, 8) = 4
    • 144 = 21 × 6 + 18, 21 = 18 × 1 + 3, 18 = 3 × 6 + 0, so the HCF is 3.
    • 3 is the 4th Fibonacci number, and 4 = HCF(12, 8).
  3. 714
    • HCF = 1, so LCM = 21 × 34 = 714.

Task C: The golden ratio

  1. φ = (1 + √5)/2 = 1.618
    • x2 − x − 1 = 0 gives x = (1 ± √5)/2; the positive root is (1 + √5)/2.
    • (1 + 2.236)/2 = 3.236/2 = 1.618
  2. Both are (3 + √5)/2
    • φ2 = (1 + 2√5 + 5)/4 = (6 + 2√5)/4 = (3 + √5)/2
    • φ + 1 = (1 + √5 + 2)/2 = (3 + √5)/2
  3. 1.619 and 1.618: very close to φ
    • 34 ÷ 21 = 1.6190…
    • 55 ÷ 34 = 1.6176…, which is 1.618 to 3 d.p.

Extension

  1. 89
    • Your last move is 1 step or 2 steps, so ways(n) = ways(n − 1) + ways(n − 2).
    • 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
  2. 34
    • The left end is one upright domino (leaving 2 by 7) or two flat dominoes (leaving 2 by 6), so ways(n) = ways(n − 1) + ways(n − 2).
    • 1, 2, 3, 5, 8, 13, 21, 34 for n = 1 to 8.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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