Express 7000 as a product of its prime factors, giving your answer in index form. Answer ___ Two integers, A and B, can be written as products of prime factors. $A = p \times q^{2} \times 7$ $B = p^{3} \times q^{r+1} \times 7$ The lowest common multiple (LCM) of A and B is 7000. Write down the value of p, q and r. Answer p = ___ q = ___ r = ___ Explain whether $H = 5^{2n+1} \times A$ is divisible by 125, where n is a whole number. Answer

1. Break 7000 into prime factors.

$7000=7\times 1000$
$1000=10^3=(2\times 5)^3=2^3\times 5^3$

2. Write the product of prime factors in index form.

$7000=7\times 2^3\times 5^3$
Hence,
$7000=2^3\times 5^3\times 7$

1. Write the LCM of $A$ and $B$ in terms of $p$, $q$ and $r$ by taking the greater power of each prime factor.

$A=p\times q^2\times 7$ and $B=p^3\times q^{r+1}\times 7$.

So
$\text{LCM}(A,B)=p^3\times q^{\max(2,r+1)}\times 7.$

2. Compare with $7000=2^3\times 5^3\times 7$ to identify $p$, $q$ and $r$.

From part (a),
$7000=2^3\times 5^3\times 7.$

Comparing
$p^3\times q^{\max(2,r+1)}\times 7=2^3\times 5^3\times 7,$
we get
$p=2,\quad q=5.$

For the power of $q$,
$\max(2,r+1)=3.$
Since $r+1$ must give the highest power $3$,
$r+1=3$
$r=2.$

1. Substitute the values of $p$ and $q$ into $A$ and express $H$ as a product of prime factors.

From part (b)(i), $p=2$ and $q=5$.

So
$A=p\times q^2\times 7=2\times 5^2\times 7$.

Hence
$H=5^{2n+1}\times A=5^{2n+1}\times 2\times 5^2\times 7=2\times 7\times 5^{2n+3}.$

2. Compare the power of 5 in $H$ with $125=5^3$ and justify divisibility.

Since $n$ is a whole number, the smallest possible value of $2n+3$ is $3$.

So $H$ contains a factor of at least $5^3$.

Since $125=5^3$, $H$ is divisible by $125$.