Primes, factor trees, HCF and LCM

(a) $1, 2, 3, 6, 9, 18$

1. Find the factor pairs: $1 \times 18$, $2 \times 9$, $3 \times 6$.

• B1 All six factors and no others.

(b) $23$ and $29$

1. $21 = 3 \times 7$, $25 = 5 \times 5$ and $27 = 3 \times 9$ have other factors. $23$ and $29$ have exactly two factors, 1 and themselves.

• B1 $23$ and $29$ only.

(c) $2^2 \times 3 \times 5$

1. Split into factors until every branch ends in a prime.
   $$60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5$$
2. Written with a power: $60 = 2^2 \times 3 \times 5$.

• M1 A correct first split, such as $2 \times 30$ or $6 \times 10$, continued at least once.
• A1 $2 \times 2 \times 3 \times 5$ or $2^2 \times 3 \times 5$.

(a) $14$

1. $$42=2\times 3\times 7$$
2. $$70=2\times 5\times 7$$
3. 42 = 2 $\times$ 3 $\times$ 7 and 70 = 2 $\times$ 5 $\times$ 7.
4. The highest common factor uses the shared prime factors: 2 $\times$ 7 = 14.

• M1 Establishing $42=2\times 3\times 7$ or an equivalent valid method.
• M1 Establishing $70=2\times 5\times 7$ or an equivalent valid method.
• A1 Correct answer: $14$

(a) 09:36

1. List multiples of each.
   $$12, 24, 36, 48, \dots \qquad 18, 36, 54, \dots$$
2. The lowest common multiple is $36$, so they next leave together $36$ minutes after 09:00.
3. That is 09:36.

• P1 Listing multiples of 12 and 18 (at least three of each), or writing $12 = 2^2 \times 3$ and $18 = 2 \times 3^2$.
• P1 Identifying $36$ minutes.
• A1 The correct answer, 09:36.

(b) $12$

1. The number of bags must divide both $48$ and $60$, so it is their highest common factor.
2. $$48 = 2^4 \times 3, \quad 60 = 2^2 \times 3 \times 5, \quad \text{HCF} = 2^2 \times 3 = 12$$

• M1 Finding common factors of 48 and 60, or their prime factorisations.
• A1 The correct answer, $12$.

(a) $12$

1. $$48=2^{4}\times 3$$
2. $$60=2^{2}\times 3\times 5$$
3. The number of packs must divide both 48 and 60 exactly. Find their highest common factor.
4. 48 = $2^{4}$ $\times$ 3 and 60 = $2^{2}$ $\times$ 3 $\times$ 5, so HCF = $2^{2}$ $\times$ 3 = 12.
5. Twelve packs work: each contains 4 red badges and 5 blue badges.

• P1 Establishing $48=2^{4}\times 3$ or an equivalent valid method.
• P1 Establishing $60=2^{2}\times 3\times 5$ or an equivalent valid method.
• A1 Correct answer: $12$

(a) $72$ seconds

1. $$18=2\times 3^{2}$$
2. $$24=2^{3}\times 3$$
3. The next shared flash is after the lowest common multiple of 18 and 24.
4. 18 = 2 $\times$ $3^{2}$ and 24 = $2^{3}$ $\times$ 3, so LCM = $2^{3}$ $\times$ $3^{2}$ = 72 seconds.

• P1 Establishing $18=2\times 3^{2}$ or an equivalent valid method.
• P1 Establishing $24=2^{3}\times 3$ or an equivalent valid method.
• A1 Correct answer: $72$ seconds

(a) $12$

1. The number of bunches must divide both flower counts.
   $$24=12\times 2$$
2. The greatest shared divisor also divides the second count.
   $$36=12\times 3$$
3. Therefore $12$.

• P1 The number of bunches must divide both flower counts.
• P1 The greatest shared divisor also divides the second count.
• A1 Correct answer: $12$

(b) $5$

1. Divide the combined flower count by the number of identical bunches.
   $$\frac{24+36}{12}$$
2. Therefore $5$.

• M1 Divide the combined flower count by the number of identical bunches.
• A1 Correct answer: $5$

(a) $60$

1. $$36=2^{2}\times 3^{2}$$
2. $$36n=12\times 180$$
3. 36 = $2^{2}$ $\times$ $3^{2}$, while 12 = $2^{2}$ $\times$ 3 and 180 = $2^{2}$ $\times$ $3^{2}$ $\times$ 5.
4. To give HCF 12, n has two factors of 2 and exactly one factor of 3. To give LCM 180, it must also have one factor of 5 and no extra prime factors.
5. Therefore n = $2^{2}$ $\times$ 3 $\times$ 5 = 60.

• P1 Establishing $36=2^{2}\times 3^{2}$ or an equivalent valid method.
• P1 Establishing $36n=12\times 180$ or an equivalent valid method.
• A1 Correct answer: $60$

(a) $36$ $(2^2 \times 3^2)$

1. Take each prime that appears in both, to the lower power: $2^2$ and $3^2$.
2. $$2^2 \times 3^2 = 36$$

• M1 Choosing the lower power of each shared prime, $2^2 \times 3^2$.
• A1 $36$ or $2^2 \times 3^2$.

(b) $2^3 \times 3^4 \times 5 \times 7 = 22\,680$

1. Take every prime that appears in either number, to the higher power.
2. $$2^3 \times 3^4 \times 5 \times 7 \;(= 22\,680)$$

• M1 Using the higher power of each prime and including both $5$ and $7$.
• A1 $2^3 \times 3^4 \times 5 \times 7$ or $22\,680$.

(c) $10$

1. In a square number every prime has an even power. In $A$, the power of $2$ is $3$ and the power of $5$ is $1$; the power of $3$ is already even.
2. Multiply by one more $2$ and one more $5$.
   $$k = 2 \times 5 = 10, \qquad A \times 10 = 2^4 \times 3^2 \times 5^2 = (2^2 \times 3 \times 5)^2$$

• P1 Recognising that each prime power must become even, and naming $2$ and $5$ as the primes to multiply by.
• A1 The correct answer, $10$.