The number $12250 \times \frac{p}{q}$ is a perfect cube. $p$ and $q$ are prime numbers. Find $p$ and $q$. Answer p = ... q = ... $x$ is a number between 700 and 800. The highest common factor of $x$ and 12 250 is 250. Find the smallest possible value of $x$. Answer $x$ = ...

1. Express $12250$ as a product of prime powers.

$12250=1225\times 10$
$=35^2\times 10$
$=(5\times 7)^2\times (2\times 5)$
$=2\times 5^3\times 7^2$

2. Use the condition for a perfect cube.

For a perfect cube, every prime factor must have an exponent that is a multiple of $3$.

In
$12250=2^1\times 5^3\times 7^2,$
$5^3$ is already suitable.

To make the power of $2$ suitable, divide by $2$:
$2^1\div 2=2^0.$
So, $q=2$.

To make the power of $7$ suitable, multiply by $7$:
$7^2\times 7=7^3.$
So, $p=7$.

3. Check that the resulting number is a perfect cube.

$12250\times \frac{7}{2}=2\times 5^3\times 7^2\times \frac{7}{2}$
$=5^3\times 7^3$
$=(5\times 7)^3=35^3,$
which is a perfect cube.

1. Use the HCF condition to restrict the form of $x$.

From part (a),
$12250=2\times5^3\times7^2.$
Also,
$250=2\times5^3.$
For $\operatorname{HCF}(x,12250)=250$, $x$ must be a multiple of $250$, but must not have an extra factor of $7$ in common with $12250$.

Between $700$ and $800$, the multiples of $250$ are
$750.$
Since $750=250\times3$ and $3$ has no common factor with $49$,
$\operatorname{HCF}(750,12250)=250.$