, 2025

A positive integer $n>1$ is
a perfect square exactly when the exponent on each prime factor in the
prime factorization of $n$ is
even.

We note that $2^3\times3^2$ has an
odd exponent on the prime factor 2, and an even exponent on the prime
factor $3$.

Thus, $2^3\times3^2\times j$ is a
perfect square exactly when $j$ is
equal to the product of an odd number of factors of $2$ and an even number of each additional
prime factor greater than $2$
(including possibly having no additional prime factors).

Since $j\leq20$, then the prime
factorization of $j$ must contain
either one factor of $2$ or three
factors of $2$, and $j$ cannot contain five or more factors of
$2$ since $2^5>20$.

When $j=2$, the given product, $2^3\times3^2\times2=2^4\times3^2=(2^2\times3)\times(2^2\times3)$,
is a perfect square.

When $j=2^3=8$, the given product,
$2^3\times3^2\times2^3=2^6\times3^2=(2^3\times3)\times(2^3\times3)$,
is also a perfect square.

Is it possible for the prime factorization of $j$ to contain one $2$ and an even number of another prime
factor greater than $2$?

The smallest prime number greater than $2$ is $3$, and when $j=2\times3^2=18<20$, the given product
$2^3\times3^2\times2\times3^2=2^4\times3^4=(2^2\times3^2)\times(2^2\times3^2)$,
is a perfect square.

The next smallest possible value of $j$ for which its prime factorization
contains exactly one 2 is $2\times5^2$, which is greater than $20$. (Note that $j=2\times3^4$ is also greater than $20$.)

The next smallest possible value of $j$ for which its prime factorization
contains exactly three factors of $2$
is $2^3\times3^2$, which is also
greater than $20$.

Therefore, the positive integers $j$
which satisfy the given conditions are $2$, $8$
and $18$.
Since $3600=60^2$ and $60=2^2\times3\times5$, then $3600=(2^2\times3\times5)^2=2^4\times3^2\times5^2$.

Thus, each divisor of $3600$ contains
at most the prime factors $2$, $3$ and $5$, and cannot contain any other prime
factors.

Further, the prime factorization of each divisor of $3600$ contains at most four factors of
$2$, two factors of $3$, and two factors of $5$.

Suppose that $k=2^x\times3^y\times5^z$. Then $\$20\times k=2^2\times5\times2^x\times3^y\times5^z=2^{x+2}\times3^y\times5^{z+1}\$$
is a divisor of $3600=2^4\times3^2\times5^2$ exactly when
$0\leq x\leq 2$, $0\leq y\leq 2$, and $0\leq z\leq 1$, for integers $x,y,z$.

As noted in part (a), $\$20\times k=2^{x+2}\times3^y\times5^{z+1}\$$ is a perfect square exactly
when each of the exponents, $x+2$,
$y$, and $z+1$ is even.

Since $0\leq x\leq 2$, then $x+2$ is even when $x=0$ or $x=2$.

Since $0\leq y\leq 2$, then $y$ is even when $y=0$ or $y=2$.

Since $0\leq z\leq 1$, then $z+1$ is even when $z=1$.

There are $2$ choices for $x$, $2$
choices for $y$, and 1 choice for
$z$, and so there are $2\times2\times1=4$ possible values of
$k$.

When $(x,y,z)=(0,0,1)$, we get $k=2^0\times3^0\times5^1=5$.

When $(x,y,z)=(0,2,1)$, we get $k=2^0\times3^2\times5^1=45$.

When $(x,y,z)=(2,0,1)$, we get $k=2^2\times3^0\times5^1=20$.

When $(x,y,z)=(2,2,1)$, we get $k=2^2\times3^2\times5^1=180$.

The positive integers $k$ satisfying
the given conditions are $5$, $45$, $20$, and $180$.
Since $2025=45^2$ and $45=3^2\times5$, then $2025=(3^2\times5)^2=3^4\times5^2$.

Since $a^2$ and $b^2$ are perfect squares, and $a^2\times b^2\times c=2025$, then $a^2$ and $b^2$ are each equal to perfect square
divisors of $2025$.

The perfect square divisors of $2025=3^4\times5^2$ are: 1, $3^2$, $3^4$, $5^2$, $3^2\times5^2$, and $3^4\times5^2$.

We count the number of ordered triples of positive integers $(a,b,c)$ by considering the following
$2$ cases: $(1)$ At least one of $a^2$ or $b^2$ is equal to $1$; $(2)$ Both $a^2$ and $b^2$ are not equal to $1$.

Case 1: At least one of $a^2$ or $b^2$ is equal to $1$.

Suppose that $a^2=1$. Then $b^2$ can be equal to each of the perfect
square divisors of $2025$ previously
listed.

That is, $b^2$ can be equal to each
of the $6$ values: $1$, $3^2$, $3^4$, $5^2$, $3^2\times5^2$, and $3^4\times5^2$.

For each of these values of $b^2$,
$c=\dfrac{2025}{a^2\times b^2}$. For
example, when $a^2=1$ and $b^2=1$, then $c=\dfrac{3^4\times5^2}{1\times 1}=3^4\times5^2$, and so $(a,b,c)=(1,1,3^4\times5^2)$ is a possible
ordered triple.

When $a^2=1$ and $b^2=3^2$, then $\$c=\dfrac{3^4\times5^2}{1\times 3^2}=3^2\times5^2$$, and so$$(a,b,c)=(1,3,3^2\times5^2)\$$ is a possible
ordered triple.

Continuing in this way with $a^2=1$,
we get the following 6 ordered triples $(a,b,c)$: $$(1,1,3^4\times5^2), (1,3,3^2\times5^2), (1,3^2,5^2), (1,5,3^4), (1,3\times5,3^2), (1,3^2\times5, 1)$$ For
each of the ordered triples above, with the exception of $(1,1,3^4\times5^2)$, we get a new ordered
triple by switching $a$ and $b$.

Each of these $5$ new ordered triples
can be determined by letting $b^2=1$
and following the process above, and thus each satisfies the given
conditions.

There are a total of $5\times2+1=11$
ordered triples in this case.

Case 2: Both $a^2$ and $b^2$ are not equal to $1$.

If one of $a^2$ or $b^2$ is equal to $3^4\times5^2$, then the other must be
equal to $1$ (since $2025=3^4\times5^2$).

In Case 2, both $a^2$ and $b^2$ are not equal to $1$, and so each is also not equal to $3^4\times5^2$.

Removing these from our list of perfect square divisors, the possible
values of $a^2$ and $b^2$ that remain are: $3^2$, $3^4$, $5^2$, and $3^2\times5^2$.

If $a^2=3^2$, then $b^2$ can be equal to $3^2$ or $5^2$ or $3^2\times5^2$.

If $a^2=3^4$, then $b^2$ can be equal to $5^2$.

If $a^2=5^2$, then $b^2$ can be equal to $3^2$ or $3^4$.

And finally, if $a^2=3^2\times5^2$,
then $b^2$ can be equal to $3^2$.

In each case, the value of $c$ can be
determined as it was in Case 1.

Doing so gives the following $7$
ordered triples $(a,b,c)$: $$(3,3,5^2), (3,5,3^2), (3,3\times5,1), (3^2,5,1), (5,3,3^2), (5,3^2, 1),(3\times5,3,1)$$ The number of ordered
triples of positive integers $(a,b,c)$ so that $a^2\times b^2\times c=2025$is $11+7=18$.