, 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$384-50=334$.