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I have a question on divisiblity of numbers I provided below.

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Does anyone know how I can approach this question? I think it may involve the prime factorization.

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    You're right about prime factorization being useful. I reccomend starting from $360^9$. If $a^3$ divides that what could $a$ be. The keep on working back. – 2017-02-17
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    I do not make any progress – 2017-02-17
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    For example, the largest $a$ can be is $360^3$. In fact, $a$ can be any factor of $360^3$. How many is that? For each one, you need to figure out how many choices there are for $b,c$. There will be the same number of choices (why?). – 2017-02-17
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    Side comment: Why did you delete your Fibonacci question? I'd like to reopen it (I have a way to a solution, I believe), but I don't want to do that if you have a pertinent reason for not wanting it reopened. – 2017-02-17

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It's very weirdly written but it is saying is find all possible sets of numbers $\{a,b,a^2, c, a^3, 360^9\}$ So that each term divides the next.

Hint: The only possible prime divisors of any of the terms are $2,3$ and $5$ and each of the terms must be of the form $2^{j}3^{k}5^{m}$ where $0 \le j \le 27; 0 \le k \le 18; 0 \le m \le 9$. ... and each term divides the next.

Play wit it first. Patterns should start to appear. What's the smallest $a$ can be? What's the largest? etc.

If $a = 2^g3^i5^k$ then $b = 2^{g'}3^{h'}5^{i'}; c= 3^{\overline g}3^{\overline h}5^{\overline i}$ and $g \le g' \le 2g \le \overline g \le 3g \le 27$ and $h \le h' \le 2h \le \overline h \le 3h \le 18$ and $i \le i' \le 2i \le \overline i \le 3i \le 9$.

So $g$ can be any $0... 9$ and $h$ can be any $0... 6$ and $i$ can be any $0...3$ For each choice there are only so many choices for the $s'$ and the $\overline s$.

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    What did *I* get as an answer? I haven't done it! – 2017-02-17