Find the real numbers given their product

Question

We have four real numbers $a,b,c,d$ and their six products should satisfy $\{ab,ac,ad,bc,bd,cd\}=\{2,2.4,3,4,5,6\}$. How do we find them?

The sequence of products is not necessarily in order. For example we don't know if $ab=2$ or $ac=2.4$.

Answer 1

$a= \sqrt{1.2}$, $b= \sqrt{10/3}$, $c=\sqrt{4.8}$, $d=\sqrt{30/4}$

I take the sequence as $\{ab,ac,ad,bc,bd,cd\}$. By the way, if the sequence changes then it would vary the values over $\{a,b,c,d\}$. But the four values will always be the same.

Answer 2

$\{a,b,c,d\}=\{\sqrt{3.6},\frac{4}{\sqrt{10}},\frac{\sqrt{10}}{2},\sqrt{10}\}$

Note that $abcd=12$, so we can first assume that ($ab=2.4$ and $cd=5$), and without loss of generality assume that ($ac=3$ and $cd=4$). Now the only thing we need to check if ($ad=2$, $cb=6$) or ($bc=2$,$ad=6$). I have checked the case when ($bc=2$,$ad=6$).

Now from $ab=2.4$, $ac=3$, $bc=2$ we get $a=\sqrt{3.6}$, $b=\frac{4}{\sqrt{10}}$, $c=\frac{\sqrt{10}}{2}$. And likely $d=\sqrt{10}$ satisfies all other 3 products.

Answer 3

Up to symmetries there are two solutions.

Inspection of the data shows that the four numbers $a$, $b$, $c$, $d$ are different, and they all have the same sign. Furthermore, when $(a,b,c,d)$ is a solution, then $(-a,-b,-c,-d)$ is a solution as well. We therefore will look for solutions satisfying $$0<a<b<c<d\ .$$ This condition implies $$ab<ac<\{bc, ad\}<bd<cd\ ,$$ so that $$ab=2,\quad ac=2.4,\quad\{bc, ad\}=\{3,4\},\quad bd=5,\quad cd=6\ .\tag{1}$$

When we try $bc=3$ then $(1)$ gives $$a^2={ab\cdot ac\over bc}=1.6,\qquad d^2={bd\cdot cd\over bc}=10\ ,$$ so that $a={2\sqrt{10}\over5}$, $\>d=\sqrt{10}$. From $(1)$ we then obtain $$(a,b,c,d)=\left({2\sqrt{10}\over5}, \>{\sqrt{10}\over2}, \>{3\sqrt{10}\over5}, \>\sqrt{10}\right)=(1.26, 1.58, 1.90, 3.16)\ .$$ It is easy to check that this quadruple fulfills all constraints.

When we try $bc=4$ in $(1)$ then we obtain $$a^2={ab\cdot ac\over bc}=1.2,\qquad d^2={bd\cdot cd\over bc}=7.5\ ,$$ so that $a={\sqrt{30}\over5}$, $\>d={\sqrt{30}\over2}$. From $(1)$ we then obtain $$(a,b,c,d)=\biggl({\sqrt{30}\over5}, \>{\sqrt{30}\over3}, \>{2\sqrt{30}\over5}, \>{\sqrt{30}\over2}\biggr)=(1.10, 1.83, 2.19, 2.74)\ .$$ It is easy to check that this is a second solution.