# On finding triplets that satisfy a certain GCD and LCM property.

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!

• Do (10,10,100), (100,10,10) count as distinct solutions? – cosmo5 Oct 20 '20 at 13:50
• @cosmo5 Yes, because $(a,b,c) = (10,10,100)$ and$(a,b,c) = (100,10,10)$ are different from each other. – Book Of Flames Oct 20 '20 at 13:51

## 2 Answers

$$10=2^15^1$$ 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.

• 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. – Book Of Flames Oct 20 '20 at 13:49
• @BookOfFlames The third exponent will be identical with one of the exponents we previously assigned, so we divide by two. – Parcly Taxel Oct 20 '20 at 13:50
• 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. – Steven Stadnicki Oct 21 '20 at 1:21

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.

• 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) – Book Of Flames Oct 21 '20 at 0:46