Written as a product of its prime factors, 360 = $2^3 \times 3^2 \times 5$. Find the prime factors of 756, giving your answer in index notation. Answer (a) (i) ... Find the highest common factor of 360 and 756. Answer (a) (ii) ... Written as a product of its prime factors, 9801 = $3^4 \times 11^2$. The number $9801^{\frac{m}{n}}$ is a perfect cube where $m$ and $n$ are prime numbers.
Find the values of $m$ and $n. Answer (b) m = ... , n = ...

1. Divide 756 repeatedly by prime factors until only prime factors remain.

$756 \div 2 = 378$

$378 \div 2 = 189$

$189 \div 3 = 63$

$63 \div 3 = 21$

$21 \div 3 = 7$

$7$ is prime.

2. Write the prime factorisation in index notation.

So

$756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7$

$756 = 2^2 \times 3^3 \times 7$

1. Compare the prime factorisations and multiply the common prime factors with the smaller indices.

From the question, $360=2^3\times3^2\times5$.

From part (a)(i), $756=2^2\times3^3\times7$.

So the HCF is
$2^2\times3^2=4\times9=36.$

1. Apply the fractional power to the prime factorisation of 9801.

$9801^{\frac{m}{n}} = (3^4 \times 11^2)^{\frac{m}{n}} = 3^{\frac{4m}{n}} \times 11^{\frac{2m}{n}}$
For this to be a perfect cube, the exponents $\dfrac{4m}{n}$ and $\dfrac{2m}{n}$ must be integers that are multiples of $3$.

2. Use the condition that $m$ and $n$ are prime numbers to find their values.

Since $n$ must divide both $4m$ and $2m$, and $m$ and $n$ are prime, take $n=2$.

Then
$\frac{4m}{2}=2m, \qquad \frac{2m}{2}=m.$
Both exponents must be multiples of $3$, so $m=3$.

Thus
$9801^{\frac{3}{2}} = 3^6 \times 11^3,$
which is a perfect cube.