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What is the number of ordered triplets $(x, y, z)$ such that the LCM of $x, y$ and $z$ is $2^33^3$ where $x, y,z\in \Bbb N$?

What I tried :

At least one of $x, y$ and $z$ should have factor $2^3$ and at least one should have factor $3^3$. I then tried to figure out the possible combinations but couldn't get the correct answer.

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We use Inclusion/Exclusion.

First we find the number of (positive) triples in which each entry divides $2^33^3$. At each of $x$, $y$, $z$ we have $(4)(4)$ choices, for a total of $16^3$.

We want to subtract the number of such triples in which each entry divides $2^23^3$. There are $12^3$ such triples. There are also $12^3$ such triples in which each element divides $2^33^2$.

But we have subtracted once too many times the $9^3$ triples in which each entry divides $2^23^2$.

So the total is $16^3-2\cdot 12^3+9^3$.

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Consider all candidate triples of the form: $$ (2^{a_1}3^{b_1}, 2^{a_2}3^{b_2}, 2^{a_3}3^{b_3}) $$ where for each $i \in \{1, 2, 3\}$, we have $a_i, b_i \in \{0, 1, 2, 3\}$.

We define such a candidate triple to be valid if for some $j, k \in \{1, 2, 3\}$, we have $a_j = 3$ and $b_k = 3$. Otherwise, if ($a_j \in \{0, 1, 2\}$ for all $j \in \{1, 2, 3\}$) or ($b_k \in \{0, 1, 2\}$ for all $k \in \{1, 2, 3\}$), then such a candidate triple is considered invalid.

Observe that: \begin{align*} \text{# of valid triples} &= \text{# of candidate triples} - \text{# of invalid triples} \\ &= 4^6 - (3^3 \cdot 4^3 + 4^3 \cdot 3^3 - 3^6) \end{align*}

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