Number Systems, Patterns and Logic — The Language of Mathematics — Practice

Question 1

(a) Find the prime factorisation of 360 and 252. Hence find HCF(360, 252) and LCM(360, 252). Verify your answer. (b) Use your LCM to add the fractions 7/360 + 5/252.

Accepted Answer

(a) PRIME FACTORISATION:
360 = 2 × 180 = 2 × 2 × 90 = 2² × 45 = 2² × 9 × 5 = 2³ × 3² × 5.
252 = 2 × 126 = 2 × 2 × 63 = 2² × 63 = 2² × 9 × 7 = 2² × 3² × 7.

HCF = product of LOWEST powers of COMMON prime factors.
Common primes: 2 (min power: 2²) and 3 (min power: 3²).
HCF(360, 252) = 2² × 3² = 4 × 9 = 36.

LCM = product of HIGHEST powers of ALL prime factors.
All primes present: 2 (max: 2³), 3 (max: 3²), 5 (max: 5¹), 7 (max: 7¹).
LCM(360, 252) = 2³ × 3² × 5 × 7 = 8 × 9 × 5 × 7 = 2520.

VERIFY: HCF × LCM = 36 × 2520 = 90,720. Also: 360 × 252 = 90,720. ✓

(b) ADDING FRACTIONS:
7/360 + 5/252. Common denominator = LCM = 2520.
7/360 = 7 × 7 / 2520 = 49/2520. (2520 ÷ 360 = 7)
5/252 = 5 × 10 / 2520 = 50/2520. (2520 ÷ 252 = 10)
49/2520 + 50/2520 = 99/2520.
Simplify: HCF(99, 2520). 99 = 9 × 11. 2520 = 2³ × 3² × 5 × 7. Common factor = 9. 99/9 = 11. 2520/9 = 280.
Final answer: 11/280.

Question 2

Write the first 6 rows of Pascal's Triangle (rows 0–5). (a) Find the sum of each row. What pattern do you notice? (b) Where do the triangular numbers appear? (c) Predict the sum of row 10 without writing it out.

Accepted Answer

PASCAL'S TRIANGLE (rows 0–5):
Row 0:          1          → sum = 1
Row 1:         1  1        → sum = 2
Row 2:        1  2  1      → sum = 4
Row 3:       1  3  3  1   → sum = 8
Row 4:      1  4  6  4  1  → sum = 16
Row 5:    1  5  10  10  5  1 → sum = 32

(a) ROW SUM PATTERN:
Sums: 1, 2, 4, 8, 16, 32 = 2⁰, 2¹, 2², 2³, 2⁴, 2⁵ = POWERS OF 2.
Rule: Sum of row n = 2ⁿ.
Why? Each element in row (n−1) contributes to EXACTLY TWO elements in row n (once left-shifted, once right-shifted). So the total DOUBLES each row.

(b) TRIANGULAR NUMBERS:
They appear in the THIRD DIAGONAL (from either side): 1, 3, 6, 10, 15, 21...
In rows 2, 3, 4, 5, 6, 7: the second number from the end of each row (reading the diagonal): 1, 3, 6, 10, 15, 21.
This is the sequence Tₙ = n(n+1)/2.

(c) PREDICTION:
Sum of row 10 = 2¹⁰ = 1024.

Question 3

Evaluate: 5 + 3 × (8 − 2)² ÷ 4 − 1. Show every step. Then explain how the Fibonacci sequence connects to the Golden Ratio, and give ONE real-world example where the Golden Ratio appears.

Accepted Answer

BODMAS EVALUATION:
Expression: 5 + 3 × (8 − 2)² ÷ 4 − 1

Step 1 — BRACKETS: (8 − 2) = 6.
→ 5 + 3 × 6² ÷ 4 − 1

Step 2 — ORDERS (exponents): 6² = 36.
→ 5 + 3 × 36 ÷ 4 − 1

Step 3 — MULTIPLICATION AND DIVISION (left to right): 3 × 36 = 108; then 108 ÷ 4 = 27.
→ 5 + 27 − 1

Step 4 — ADDITION AND SUBTRACTION (left to right): 5 + 27 = 32; then 32 − 1 = 31.

FINAL ANSWER: 31.

FIBONACCI AND THE GOLDEN RATIO:
The Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89...

Consider the RATIO of consecutive terms:
2/1 = 2.000
3/2 = 1.500
5/3 = 1.667
8/5 = 1.600
13/8 = 1.625
21/13 = 1.615
34/21 = 1.619
55/34 = 1.618...

As n → ∞, the ratio F(n+1)/F(n) → φ ≈ 1.6180339... (Golden Ratio).

The Golden Ratio φ satisfies: φ² = φ + 1 (i.e. φ = (1+√5)/2).

REAL-WORLD EXAMPLE: Sunflower seed heads — the number of clockwise and anti-clockwise spirals are ALWAYS consecutive Fibonacci numbers (typically 34 and 55, or 55 and 89). This mathematical pattern minimises wasted space and maximises seed packing efficiency — it evolved through natural selection because it is optimal. The Fibonacci/Golden Ratio pattern also appears in pinecones (8 and 13 spirals), pineapples, and the shell of the nautilus.

Practice — Number Systems, Patterns and Logic — The Language of Mathematics

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