3 (a) (i) Write 120 as a product of its prime factors - OCR - GCSE Maths - Question 3 - 2018 - Paper 1

Given that the LCM of x and 120 is 360, we first find the prime factorization of 360:

1. Divide by 2: 360 ÷ 2 = 180
2. Divide by 2: 180 ÷ 2 = 90
3. Divide by 2: 90 ÷ 2 = 45
4. Divide by 3: 45 ÷ 3 = 15
5. Divide by 3: 15 ÷ 3 = 5
6. Divide by 5: 5 ÷ 5 = 1.

Thus, 360 can be expressed as:

$360 = 2^3 × 3^2 × 5^1$

We note that 120 is:

$120 = 2^3 × 3^1 × 5^1$

To find x, we analyze the LCM condition. The LCM takes the highest power of each prime factor from the numbers involved. Therefore:

• For 2: both have $2^3$
• For 3: $x$ must have at least $3^2$ to achieve the LCM of $3^2$.
• For 5: both have $5^1$

Thus, we can express x as:

$x = 2^0 × 3^2 × 5^0 = 9$.

To find the HCF of A and B, we will take the lowest power of each common prime factor:

• For 2: The powers are $2^4$ in A and $2^3$ in B. So, we take $2^3$.
• For 3: The powers are $3^2$ in A and $3^1$ in B. So, we take $3^1$.
• For 7: The powers are $7^2$ in A and $7^1$ in B. So, we take $7^1$.
• 5 is not a common factor.
  Thus, the HCF can be calculated as:

$HCF(A,B) = 2^3 × 3^1 × 7^1 = 8 × 3 × 7 = 168$.