Express 1400 as the product of its prime factors. Answer ... Write down the smallest positive integer k such that $1400k$ is a perfect cube. Answer k = ... n is a number between 300 and 400. The highest common factor of n and 1400 is 35. Find the largest possible value for n. Answer n = ...

1. Find the prime factorisation of 1400.

$1400 = 14 \times 100$
$= (2 \times 7)(2^2 \times 5^2)$
$= 2^3 \times 5^2 \times 7$

1. Use prime factorisation and make all indices multiples of 3.

From part (a),

$1400=2^3\times 5^2\times 7^1$.

For $1400k$ to be a perfect cube, the powers of all prime factors must be multiples of $3$.

$2^3$ is already a cube factor.

$5^2$ needs one more $5$, and $7^1$ needs two more $7$'s.

So

$k=5\times 7^2=245$.

1. Use the HCF condition to express $n$ as a multiple of 35 and find the possible range of the multiplier.

Since the highest common factor of $n$ and $1400$ is $35$, $n$ must be a multiple of $35$.

Let $n=35m$.

Given $300<n<400$,
$300<35m<400$
$8.57\ldots<m<11.42\ldots$
So $m=9,10,$ or $11$.

2. Choose the largest multiple of 35 in the range whose HCF with 1400 is exactly 35.

The possible values of $n$ are
$35\times 9=315,\quad 35\times 10=350,\quad 35\times 11=385.$

Check the largest first:
$1400=35\times 40,$
$385=35\times 11.$
Since $11$ has no common factor with $40$, the HCF is exactly $35$.

Therefore the largest possible value is
$n=385.$