Solutions: Section 5.4

Page 120: Figure it Out

Q1. Prime Factorisations

• 64: $2 \times 2 \times 2 \times 2 \times 2 \times 2$ ($2^6$)
• 105: $3 \times 5 \times 7$
• 320: $32 \times 10 = 2^5 \times 2 \times 5 = 2^6 \times 5$ ($2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$)
• 1000: $10 \times 10 \times 10 = (2 \times 5)^3 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$

Q2. Identify the number

Factors: one 2, two 3s, one 11. $\text{Number} = 2 \times (3 \times 3) \times 11$ $= 2 \times 9 \times 11$ $= 18 \times 11 = \mathbf{198}$

Q3. Three primes < 30 product 1955

Ends in 5, so 5 is a factor. $1955 \div 5 = 391$. We need two primes that multiply to 391. Try primes: 7 (no), 11 (no), 13 (no), 17. $391 \div 17 = 23$. 23 is prime. Answer: 5, 17, 23.

Page 122: Figure it Out

Q1. Co-prime check via factorisation

a. 30 and 45 $30 = 2 \times 3 \times 5$ $45 = 3 \times 3 \times 5$ Common factors 3 and 5. No.

b. 57 and 85 $57 = 3 \times 19$ $85 = 5 \times 17$ No common factors. Yes.

Q2. Divisibility check via factorisation

a. 225 divisible by 27? $225 = 5 \times 5 \times 3 \times 3$ $27 = 3 \times 3 \times 3$ 225 only has two 3s. 27 needs three 3s. No.

b. 96 divisible by 24? $24 = 2 \times 2 \times 2 \times 3$ $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3$ 96 contains all factors of 24. Yes.