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The question is this:

Find all triplets of positive integers $a,b,c$ satisfying $(a,b,c) = 10$ and $\left[a,b,c\right] = 100$ simultaneously. Here, $(x,y)$ is the greatest common divisor of $x$ and $y$ and $[x,y]$ is the least common multiple of $x$ and $y$.

It would be very easy if the question was in $2$ variables, because there is a relation between product of the numbers, the GCD and the LCM, but I am not aware of any relation between them when $3$ variables are involved.

For attempting purpose, one can set $a = 10p$, $b = 10q$ and $c = 10r$, where $(p,q,r) = 1$.
Then $[p,q,r] = 10$. Now I don't have an idea to proceed from here.
I am pretty much a beginner to elementary number theory, so might be I have missed something obvious.

Thanks for solutions!

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  • $\begingroup$ Do (10,10,100), (100,10,10) count as distinct solutions? $\endgroup$ – cosmo5 Oct 20 '20 at 13:50
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    $\begingroup$ @cosmo5 Yes, because $(a,b,c) = (10,10,100)$ and$(a,b,c) = (100,10,10)$ are different from each other. $\endgroup$ – Book Of Flames Oct 20 '20 at 13:51
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$10=2^15^1$ and $100=2^25^2$. Let $a=2^{a_2}5^{a_5}$, and similarly for $b$ and $c$. Because the gcd expresses the minimum exponent for each prime across its arguments, we have $$\min(a_2,b_2,c_2)=1\qquad \min(a_5,b_5,c_5)=1$$ Similarly, the lcm expresses the maximum prime exponents, and we have $$\max(a_2,b_2,c_2)=2\qquad \max(a_5,b_5,c_5)=2$$ One of $a_2,b_2,c_2$ has to be $1$ and another $2$. The third may be either $1$ or $2$. This gives six possibilities for the triple $(a_2,b_2,c_2)$ (as can be directly enumerated easily), and by symmetry there are six possibilities for $(a_5,b_5,c_5)$ independent of the other set of variables. Thus there are $6×6=36$ triples $(a,b,c)$ satisfying the conditions.

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  • $\begingroup$ Can you please explain that how if the third was either of $1$ or $2$, then the number of possibilities is $6$? Due to symmetry we get $6$, then because the third can be either of $1$ or $2$ ... then it sums to $12$, I guess. $\endgroup$ – Book Of Flames Oct 20 '20 at 13:49
  • $\begingroup$ @BookOfFlames The third exponent will be identical with one of the exponents we previously assigned, so we divide by two. $\endgroup$ – Parcly Taxel Oct 20 '20 at 13:50
  • $\begingroup$ Another way of looking at it: for the exponents of 2 for instance (i.e., $a_2$, $b_2$, $c_2$) there's either one 1 and two 2s, or two 1s and one 2. In either case, there are three possibilities, making six total. $\endgroup$ – Steven Stadnicki Oct 21 '20 at 1:21
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All three are divisible by $10$, so we divide each by $10$ and find the triples with GCD = $1$ and LCM = $10$.

There can only be factors $2$ and $5$, and both must be there. So, the numbers can be $1$, $2$, $5$ or $10$, and either $2$ and $5$ or just $10$ must be present. They cannot all three be divisible by $2$, or all three divisible by $5$. If we take the numbers in sorted order:

$$ (1,1,10);\ (1,2,5);\ (1,2,10);\ (1,5,10);\ (1,10,10);\ (2,2,5);\ (2,5,5);\ (2,5,10) $$

That’s $8$ solutions, and you can arrange the numbers in different order. Multiply by $10$ to solve the original problem.

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  • $\begingroup$ That too is nice ... It sums to $3 + 6 + 6 + 6 + 3 + 3 + 3 + 6 = 36$. (The LHS is written in order of the number of arrangements of the triplets) $\endgroup$ – Book Of Flames Oct 21 '20 at 0:46

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