Write 756 as a product of its prime factors. Answer: ... Given that $495 = 3^2 \times 5 \times 11$, find the smallest positive integer $p$ such that $756p$ is a multiple of 495. Answer: ... The number $756 \times \frac{a}{b}$ is a perfect cube where $a$ and $b$ are both prime numbers. Find the smallest possible value of $a$ and smallest possible value of $b$. Answer: a = ... , b = ...

1. Break 756 down into prime factors.

$756 = 2 \times 378$

$378 = 2 \times 189$

$189 = 3 \times 63$

$63 = 3 \times 21$

$21 = 3 \times 7$

2. Write the prime factors in index form.

So,

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

1. Use the prime factorisation of $756$ and compare it with $495$.

From part (a),
$756=2^2\times 3^3\times 7.$
Given
$495=3^2\times 5\times 11.$
For $756p$ to be a multiple of $495$, it must contain at least $3^2\times 5\times 11$ as factors.

$756$ already contains $3^3$, so the missing factors are $5$ and $11$.

$p=5\times 11=55$

1. Use the prime factorisation of 756 and identify how the exponents must change to form a perfect cube.

From part (a),
$756 = 2^2 \times 3^3 \times 7.$
For $756 \times \dfrac{a}{b}$ to be a perfect cube, each prime factor must have an exponent that is a multiple of $3$.

The exponent of $2$ is $2$, so dividing by $2$ makes it $1$, not a cube exponent, but multiplying by $2$ makes it $3$.
The exponent of $7$ is $1$, so multiplying by $7^2$ would be needed, but $a$ must be prime, so this is not possible using $a$ alone.

Instead, choose $b=2$:
$756 \div 2 = 2^1 \times 3^3 \times 7,$
then multiplying by $a=7$ gives
$756 \times \frac{7}{2}=2^1 \times 3^3 \times 7^2,$
which is not a perfect cube. So this does not work.

2. Find prime values of $a$ and $b$ that make every exponent a multiple of 3.

Write
$756 \times \frac{a}{b}=2^2 \times 3^3 \times 7 \times \frac{a}{b}.$
To make the exponent of $2$ a multiple of $3$, multiply by $2$, so take $a=2$.

To make the exponent of $7$ a multiple of $3$, divide by $7$, since $7^1 \div 7 = 7^0$ and $0$ is a multiple of $3$, so take $b=7$.

Then
$756 \times \frac{2}{7}=2^3 \times 3^3,$
which is a perfect cube.