Problem 2. It is known that $a, b$ and $c$ are natural numbers, $LCM(a, b)=945$, $LCM(b, c)=525$. What can $\operatorname{LCM}(a, c)$ be equal to?
# Solution:

Let's factorize the numbers into prime factors, since $3^{3}$ only appears in $H O K(a, b)$, it follows that $a: 3^{3}$, similarly we get that $b: 5^{2}$

$$
\left\{\begin{array} { l } 
{ \operatorname { H O K } ( a , b ) = 9 4 5 = 3 ^ { 3 } \cdot 5 \cdot 7 } \\
{ \operatorname { H O K } ( b , c ) = 5 2 5 = 3 \cdot 5 ^ { 2 } \cdot 7 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a: 3^{3} \\
b \vdots 5^{2}
\end{array} \Rightarrow \operatorname{HOK}(a, c):\left(3^{3} \cdot 5^{2}\right)\right.\right.
$$

Notice that,

$$
\begin{gathered}
\left\{\begin{array}{l}
\operatorname{HOK}(a, b, c) \vdots \operatorname{HOK}(a, b) \\
\operatorname{HOK}(a, b, c) \vdots \operatorname{HOK}(b, c) \\
\operatorname{HOK}(a, b, c) \vdots \operatorname{HOK}(a, c)
\end{array}\right. \\
\Rightarrow \operatorname{HOK}(a, b, c) \vdots \operatorname{HOK}(945,525) \Rightarrow \operatorname{HOK}(a, b, c):\left(3^{3} \cdot 5^{2} \cdot 7\right)
\end{gathered}
$$

Considering three facts: $\left\{\begin{array}{c}\operatorname{HOK}(a, b, c):\left(3^{3} \cdot 5^{2} \cdot 7\right) \\ H O K(a, b, c) \vdots H O K(a, c), \text { we get that } \\ \operatorname{HOK}(a, c):\left(3^{3} \cdot 5^{2}\right)\end{array}\right.$

$\operatorname{HOK}(a, c)=3^{3} \cdot 5^{2}$, for example, when $a=3^{3} \cdot 5, b=7, c=3 \cdot 5^{2}$ or

LCM $(a, c)=3^{3} \cdot 5^{2} \cdot 7$, for example, when $a=3^{3} \cdot 5, b=7, c=3 \cdot 5^{2} \cdot 7$

Answer: 675 or 4725