, 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$,$$c=\dfrac{2025}{a^2\times}$ b^2}$. For example, when$a^2=1$and$b^2=1$\text{, then (c= 3 4 5 2 1 1 =3 4 5 2 ), and so}$$(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$11+7=18$.