Foundation June 2025 Paper 2R Q18
18 Write 1400 as a product of powers of its prime factors.
Show your working clearly.
(3)
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| eg \(2 \times 2 \times 350\) or \(2 \times 7 \times 100\) or \(2 \times 5 \times 140\) or \(5 \times 5 \times 56\) or \(7 \times 5 \times 40\) or \((14 \times 100 = 14 \times 25 \times 4 =)\;2 \times 7 \times 25 \times 4\) eg
or eg  | M1 | ||||||||||||
| eg \(2 \times 2 \times 2 \times 5 \times 5 \times 7\) eg
or eg  | M1 | ||||||||||||
| Working required Answer: \(2^3 \times 5^2 \times 7\) | A1 | ||||||||||||
| (3) | |||||||||||||
| (3 marks) |
Notes
M1: for finding 2 prime factors after at least 2 stages of prime factorisation with 0 incorrect stages
or for finding 2 prime factors after at least 3 stages of prime factorisation with no more than 1 incorrect stage
Each stage gives 2 factors – may be in a factor tree or a table or listed (see LHS for examples of the amount of work needed for the award of this mark) but we want to see 2 prime factors.
Example of finding 2 prime factors after at least 3 stages with 1 incorrect stage:
1400 = 10 × 14 = 2 × 5 × 2 × 7
M1: dep on M1
for factors 2, 2, 2, 5, 5, 7 identified with no others in any form, eg listed, multiplied, added
Ignore 1s
May be seen in a fully correct factor tree or ladder
A1: dep on M2
May be in any order and allow \(2^3 \text{x} 5^2 \text{x} 7\)