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I tried to solve this problem but I was not able to. Can someone please tell me the way to solve this problem in simple ways? I have looked up for the solution to this problem on the web but I wasn't able to get a well explained answer. I am just looking for unordered pairs of solutions.

One more question related to this problem. So in this problem we have been asked to find number of pairs of $(a,b)$ such that $LCM(a,b)=200=2^3 \times 5^2$ but is there a generalized way to solve for triplets, quadruplets etc.too? Like how many triplets of $(a,b,c)$ such that $LCM(a,b,c)=200=2^3 \times 5^2$ or how many quadruplets of $(a,b,c,d)$ such that $LCM(a,b,c,d)=200=2^3 \times 5^2$

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  • $\begingroup$ Why not start with a smaller number, for which you can just list all the options? That should show you the pattern. It's not subtle. Note, you should indicate whether you mean ordered or unordered pairs. $\endgroup$
    – lulu
    Oct 1, 2021 at 11:23
  • $\begingroup$ can you commennt on the prime factors of $a,b,c,d$ $\endgroup$ Oct 1, 2021 at 11:24
  • $\begingroup$ @lulu : Ok so I checked for LCM=20. the pairs are (1,20) (20,20) (4,5) (10,4) (20,2) (5,20) (10,20) (20,4). I hope I have not missed any. The pattern that I am seeing here is the factors of 20 are being multiplied with each other. $\endgroup$
    – Ganit
    Oct 1, 2021 at 11:32

1 Answer 1

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Hint, consider an ordered pair $(X, Y)$. Write \begin{align} X &= P_1^{a_1}P_2^{a_2}P_3^{a_3}\ldots P_n^{a_n} \\ Y &= P_1^{b_1}P_2^{b_2}P_3^{b_3}\ldots P_n^{b_n} \end{align} Here, $P_1, P_2,\ldots P_n$ are prime factors of $X$ and $Y$.

Now $$\text{LCM}(X, Y) = P_1^{max(a_1, b_1)}P_2^{max(a_2, b_2)}\ldots P_n^{max(a_n, b_n)}$$

Therefore $$\text{LCM}(X, Y) = N = P_1^{m_1}P_2^{m_2}P_3^{m_3}\ldots P_n^{mn}$$

Where $m_i=\text{max}(a_i,b_i)$. Therefore, total number of ordered pairs \begin{align} (X, Y) &= ((m_1 + 1)^2 – m_1^2))((m_2 + 1)^2 – m_2^2)\ldots((m_n + 1)^2 – m_n^2)\\ &= (2m_1+1) (2m_2+1)(2m_3+1)\ldots(2m_n+1) \end{align}

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  • $\begingroup$ I didn't get the last part where you calculated the total number of ordered pairs. Also this must be counting one pair twice like (1,20) and (20,1). So in order to find the unordered pair i need to do (n+1)/2,right? also how about finding triplets and quadruplets? $\endgroup$
    – Ganit
    Oct 1, 2021 at 11:36
  • $\begingroup$ There were a few typos - hopefully that's cleared it up? $\endgroup$
    – user284001
    Oct 1, 2021 at 11:59

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