Prime factors
Number
- Foundation
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
-
Write $84$ as a product of its prime factors.
2 marks
-
Write $180$ as a product of its prime factors.
2 marks
-
Show that $91$ is not a prime number. Is $97$ a prime number?
3 marks
-
Use prime factorisations to find the HCF of $60$ and $84$.
3 marks
-
Use prime factorisations to find the LCM of $60$ and $84$.
4 marks
-
Find the smallest positive integer $n$ such that $180n$ is a square number.
4 marks
-
$a = 2^{3} \times 3^{2} \times 5$ and $b = 2 \times 3 \times 5^{2}$. Find the HCF of $a$ and $b$, and the LCM of $a$ and $b$.
5 marks
-
$p = 2^{2} \times 3 \times 7$ and $q = 2 \times 3^{2} \times 5$. Find the HCF and LCM of $p$ and $q$. Then find the smallest positive integer $k$ such that $pk$ is a square number, and write $pk$ as a square.
6 marks
Mark scheme
-
Answer. $2^{2} \times 3 \times 7$
- $84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7$.
- $2^{2} \times 3 \times 7$
-
Answer. $2^{2} \times 3^{2} \times 5$
- $180 = 18 \times 10 = 2 \times 3^{2} \times 2 \times 5$.
- $2^{2} \times 3^{2} \times 5$
-
Answer. $91 = 7 \times 13$, so it is not prime. $97$ is prime.
- $91 = 7 \times 13$.
- $97$ is not divisible by $2, 3, 5$ or $7$.
- The next prime is $11$, and $11 \times 9 = 99$, so $97$ is prime.
-
Answer. $12$
- $60 = 2^{2} \times 3 \times 5$ and $84 = 2^{2} \times 3 \times 7$.
- HCF $= 2^{2} \times 3 = 12$.
-
Answer. $420$
- $60 = 2^{2} \times 3 \times 5$ and $84 = 2^{2} \times 3 \times 7$.
- LCM $= 2^{2} \times 3 \times 5 \times 7$.
- $4 \times 3 \times 5 \times 7 = 420$.
-
Answer. $5$
- $180 = 2^{2} \times 3^{2} \times 5$.
- A square needs an even power of every prime, so one more factor of $5$ is needed.
- $n = 5$, and $180 \times 5 = 900 = 30^{2}$.
-
Answer. HCF $30$, LCM $1800$
- HCF $= 2 \times 3 \times 5 = 30$.
- LCM $= 2^{3} \times 3^{2} \times 5^{2}$.
- $8 \times 9 \times 25 = 1800$.
-
Answer. HCF $6$, LCM $1260$, $k = 21$, $pk = 42^{2}$
- HCF $= 2 \times 3 = 6$.
- LCM $= 2^{2} \times 3^{2} \times 5 \times 7 = 1260$.
- $p = 2^{2} \times 3 \times 7$, so $k = 3 \times 7 = 21$.
- $pk = 2^{2} \times 3^{2} \times 7^{2} = 42^{2}$.