Number 3

5 marks · Hard

Let N = 2940.

(a)

Express 2940 as a product of its prime factors in index form.

[2 marks]

(b)

Given that M = 2940 × 42, write M as a product of its prime factors in index form.

[2 marks]

(c)

Hence or otherwise, find the highest common factor (HCF) of 2940 and 42.

[1 mark]

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Show answer

(a) 2940 = 2^2 × 3 × 5 × 7^2

(b) M = 2^3 × 3^2 × 5 × 7^2

Common mistakes: missing factors in part a; not multiplying exponents in part b; using LCM instead of HCF in c

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Correct attempt to break down 2940 into primes.

A1 (a) · 1 mark · Express 2940 accurately as 2^2 × 3 × 5 × 7^2.

M2 (b) · 1 mark · Uses correct method for multiplication of prime factor powers. (follow-through)

A2 (b) · 1 mark · Obtains correct index form for M. (follow-through)

A3 (c) · 1 mark · Finds HCF = 42 (2 × 3 × 7) from factorisations. (follow-through)

Worked solution

a. 2940:

2940 ÷ 2 = 1470, ÷ 2 = 735, ÷ 3 = 245, ÷ 5 = 49, ÷ 7 = 7, ÷ 7 = 1.

So 2940 = 2^2 × 3 × 5 × 7^2.

b. 42 = 2 × 3 × 7.

M = 2940 × 42 = (2^2 × 3 × 5 × 7^2) × (2 × 3 × 7) = 2^3 × 3^2 × 5 × 7^3.

(Correct exponents: 2's: 2+1=3; 3's: 1+1=2; 5's: 1; 7's: 2+1=3)

c. Common factors of 2940 and 42: 2 × 3 × 7 = 42, so HCF = 42.