We have 4 positive,but not necessarily integers $a, b, c,$ and $d$. So we have 6 options how multiply two of them. And we know 5 of 6 products $2, 3, 4, 5$ and $6$. Find the last product.

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    $\begingroup$ The answer is $\frac{12}{5}$, as LaBird demonstrates. But there are two possibilities for $\{a,b,c,d\}$: $\{\sqrt{\frac52}, \sqrt{\frac85}, \sqrt{\frac{18}5},\sqrt{10}\}$ and $\{\sqrt{\frac65},\sqrt{\frac{10}{3}},\sqrt{\frac{15}{2}}, \sqrt{\frac{24}{5}}\}$. $\endgroup$
    – TonyK
    Mar 26 '15 at 14:52

One way is to find out which two pairs of the $5$ known numbers give the same product, and then use this product to divide the remaining known number. In your example, by inspection, $2 \times 6$ = $3 \times 4$ = $12$, then $12 / 5 = 2.4$, so the last product is $2.4$.

The reason is: We assume the one pair of product is $(a \times b) \times (c \times d)$, the other pair of product is $(a \times c) \times (b \times d)$, and $(a \times b) \times (c \times d) = (a \times c) \times (b \times d) = (a \times d) \times (b \times c)$, so if we find two pairs of the $5$ known numbers can give the same product, it will be the product of $abcd$. Dividing it by the remaining number $(a \times d)$ gives the remaining product $(b \times c)$.

Edit: This method may sometimes give more than one solution. Consider the case when the given $5$ products are $8, 12, 16, 24$ and $32$, you may pair up $8 \times 24 = 12 \times 16 = 192$, and then divide by $32$ to get $6$ (the $4$ original numbers are $2, 3, 4, 8$ or $\sqrt{3}, \sqrt{12}, \sqrt{\frac{64}{3}}, \sqrt{48}$). On the other hand, you may also pair up $12 \times 32 = 16 \times 24 = 384$, and then divide by $8$ to get $48$ (the $4$ original numbers are then $2, 4, 6, 8$ or $\sqrt{6}, \sqrt{\frac{32}{3}}, \sqrt{24}, \sqrt{96}$).


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