P And C

First of all, LCM is the product of the primes with the highest power among two numbers. For example, if

$p = {{a}_{1}}^{{\alpha}_{1}} {{a}_{2}}^{{\alpha}_{2}} {{a}_{3}}^{{\alpha}_{3}}.......$

$q = {{a}_{1}}^{{\beta}_{1}} {{a}_{2}}^{{\beta}_{2}} {{a}_{3}}^{{\beta}_{3}}......$

and if ${\alpha}_{1}>{\beta}_{1}, {\alpha}_{2}>{\beta}_{2},.....$

Then, $LCM(p,q) = {{a}_{1}}^{{\alpha}_{1}} {{a}_{2}}^{{\alpha}_{2}} {{a}_{3}}^{{\alpha}_{3}}.......$

Case 1 - For getting ${2}^{2}$ either p is a multiple of ${2}^{2}$ or q

When p is a multiple of ${2}^{2}$ , then q can be a multiple of ${2}^{0}, {2}^{1}, {2}^{2}$

When q is a multiple of ${2}^{2}$ , then p can be a multiple of ${2}^{0}, {2}^{1}, {2}^{2}$

So, there are 6 pairs for getting ${2}^{2}$ as a factor of LCM(p,q). But $({2}^{2},{2}^{2}) = ({2}^{2},{2}^{2})$ , hence there are $6-1 = 5$ ordered pairs.

Similarly Case 2 and Case 3 can be solved as well yielding the answers as 9 and 5 respectively.

Now using the Law of Multiplication, there are $5*9*5 = 225$ ordered pairs.

p,q must be in the form ${ r }^{ a }{ t }^{ b }{ s }^{ c }$

where a=0,1,2 b=0,1,2,3,4 c=0,1,2

Let's write the powers of factors of p and q in a table form: $\begin{matrix} & r & t & s \\ p & a_{p} & b_{p} & c_{p} \\ q & a_{q} & b_{q} & c_{q}\end{matrix}$

Consider factor r across p and q, ie. row $\begin{matrix} r \\ a_{p} \\ a_{q}\end{matrix}$

No. of combinations without constrain $=3 \times 3=9$

However at least one of $a_{p}$ and $a_{q}$ has to be 2 to obtain LCM $r^{2}$

No. of combinations if none of $a_{p}$ and $a_{q}$ equals to 2 $= 2\times 2 = 4$

∴ No. of actual combinations of r $= 9 - 4 = 5$

Similar for factor t and s, we have:

No. of actual combinations of t $= 5^{2} - 4^{2} = 9$

No. of actual combinations of s $= 3^{2} - 2^{2} = 5$

Hence, No. of ordered pair (p,q) $= 5 \times 9 \times 5 = 225$