, 2025

The prime factorization of $400$ is $2^45^2$, and so $400$ has $(4+1)(2+1)=15$ positive divisors.

Since $20^2=400$, one of these $15$ positive divisors is $20$, from which we get the ordered pair
$(a,b)=(20,20)$.

The remaining $14$ positive divisors
give $\dfrac{14}{2}=7$ factor pairs
of positive integers $(a,b)$ with
$a< b$. One of these $7$ factor pairs has $a=1$, which we must omit since $a>1$.

Thus, there are $6$ ordered pairs
$(a,b)$ with $a<b$ and 1 with $a=b$, for a total of $7$ ordered pairs of positive integers
$(a,b)$ for which $ab=400$ and $1<a\leq b$.

(These ordered pairs are $(2,200)$,
$(4,100)$, $(5, 80)$, $(8,50)$, $(10,40)$, $(16,25)$, and $(20,20)$.)
The prime factorization of $270 000$ is $2^43^35^4$.

We count the number of ways to distribute each of the prime factors
among $p$, $q$ and $r$.

The three prime factors equal to $3$ could all be distributed to exactly one
of $p$, $q$ or $r$.

This can be done in $3$ different
ways.

Two $3$s can be distributed to one
factor, and one $3$ to another. There
are $3$ choices for the factor having
two $3$s, and $2$ choices for the factor having one $3$, and thus $3\times2=6$ ways to distribute two $3$s to one factor and one $3$ to another.

Finally, one $3$ can be distributed
to each of the factors, and this can be done in $1$ way.

Thus, there are $3+6+1=10$ ways to
distribute the $3$s among $p$, $q$
and $r$.

The four prime factors equal to $2$ could all be distributed to exactly one
of $p$, $q$ or $r$.

This can be done in $3$ different
ways.

Three $2$s can be distributed to one
factor, and one $2$ to another. There
are $3$ choices for the factor having
three $2$s, and $2$ choices for the factor having one $2$, and thus $3\times2=6$ ways to distribute three $2$s to one factor and one $2$ to another.

Two $2$s can be distributed to one
factor, and two $2$s to another
factor. There are $3$ choices for the
factor having two $2$s, and $2$ choices for the second factor having
two $2$s. However, since the number
of $2$s being distributed to each
factor is equal, we have double counted the possibilities.

Thus, there are $\dfrac{3\times2}{2}=3$ ways to distribute
two $2$s to one factor and two $2$s to another.

(This is equivalent to counting the number of ways of choosing the
factor receiving no $2$s.)

Finally, two $2$s can be distributed
to one of the factors in $3$ ways,
and one $2$ can be distributed to
each of the other two factors in $1$
way, for $3$ possibilities in this
final distribution. Thus, there are $3+6+3+3=15$ ways to distribute the $2$s among $p$, $q$
and $r$.

Similarly, there are $15$ ways to
distribute the four $5$s among $p$, $q$
and $r$, and so there are $10\times15\times15=2250$ ways to
distribute the prime factors, and thus $2250$ ordered triples of positive integers
$(p,q,r)$ for which $pqr=270\,000$.
As shown in part (b), there are $2250$ ordered triples of positive integers
$(x,y,z)$ for which $xyz=270\,000=2^43^35^4$.

For distinct values of $x$, $y$ and $z$, these $2250$ triples include the $6$ possible arrangements of $x$, $y$
and $z$.

For example, each of the $6$
arrangements of $(5^4,2^4, 3^3)$ is
included among the $2250$
triples.

Of these $6$, it is only $(2^4, 3^3, 5^4)$ that is counted in part
(c) since we require $\$x\leq y\leq$
z$.

Thus, we must determine the number of ordered triples from part (b)
which do not satisfy $x\leq y\leq z$
and subtract this from $2250$ (we
refer to this as Step 1).

Further, we require ordered triples for which $1<x$. Thus, we also must determine the
number of ordered triples for which $x=1$ and subtract this from the number of
triples that remain following Step 1. We refer to this as Step 2.

Step 1:

Since $270\,000$ is not a perfect
cube, it is not possible that $x=y=z$.

Thus, each of the $2250$ ordered
triples belongs to exactly one of the following two cases: exactly two
of $x$, $y$, $z$
are equal, or all three are distinct.

Suppose that exactly two of the factors are equal, say $x=y$.

In this case $xyz=x^2z=2^43^35^4$,
and so $x^2$ is a perfect square
divisor of $2^43^35^4$.

Each perfect square divisor of $2^43^35^4$ is a number of the form $2^u3^v5^w$ where $u=0,2$ or 4, $v=0$ or $2$, and $w=0,2$ or $4$.

There are $3$ choices for $u$, $2$
choices for $v$, and $3$ choices for $w$, and so there are $3\times2\times3=18$ perfect square
divisors of $2^43^35^4$.

When exactly two of the factors are equal, there are $3$ ways to arrange the three factors, and
so there are $3\times18=54$ ordered
triples for which exactly two of $x$,
$y$, $z$ are equal.

Thus, the number of ordered triples for which $x$, $y$
and $z$ are distinct is $2250-54=2196$.

For distinct values of $x$, $y$ and $z$, the $2196$ triples include each of the 6
possible arrangements of $x$, $y$ and $z$.

Thus, the number of ordered triples $(x,y,z)$ with $1\leq x<y<z$ is $\dfrac{2196}{6}=366$.

Since there are $18$ ordered triples
$(x,y,z)$ for which $x$, $y$, $z$
are not distinct (exactly two are equal), then there are $366+18=384$ ordered triples $(x,y,z)$ with $1\leq x \leq y \leq z$. This completes
Step 1.

Step 2:

We require each of $x$, $y$, $z$
to be greater than $1$, and so in
this final step we determine the number of ordered triples for which
$x=1$, and subtract this from $384$.

When $x=1$, $xyz=2^43^35^4$ becomes $yz=2^43^35^4$ which means that $(y,z)$ is a factor pair of $2^43^35^4$.

Since $2^43^35^4$ has $5\times4\times5=100$ positive divisors,
then it has $\dfrac{100}{2}=50$
factor pairs of positive integers $(y,z)$ with $y\leq z$, and so there are $50$ ordered triples for which $x=1$.

Thus, the number of ordered triples of positive integers $(x,y,z)$ for which $xyz=270\,000$ and $1<x\leq y \leq z$ is $384-50=334$.