## What are the Numbers?

1. #### lvps1000vm:

The thing to notice is that each product shares a factor with four other and no factors with one. So we can partition the three products in three pairs. Multiplying a product with “its pair” gives the total product of the four original numbers. One must find two such pairs that multiply to a common total.

So, 2·6 = 12 and 3·4 = 12, therefore 5·x = 12. The remaining product is 12/5.

If x is the common factor of 2 and 3, the original numbers are x, 2/x, 3/x and 2x. (They multiply to 12), they give two equations:
2x^2 = 5 and 6x^-2 = 12/5 => x = sqrt(5/2)
or
6x^-2 = 5 and 2x^2 = 12/5 => x = sqrt(6/5)

The original numbers could be {sqrt(5/2), sqrt(8/5), sqrt(18/5), sqrt(10)} or {sqrt(6/5), sqrt(10/3), sqrt(15/2), sqrt(24/5)}

2. #### Oscar Cunningham:

Let the two numbers used to make the product we haven’t been given be a and b, and let the others be c and d. Then ac, ad, bc, bd and cd are 2, 3, 4, 5 and 6 in some order. Since (ac)(bd) = (ad)(bc) = abcd we must have that two pairs of our products can themselves be multiplied to give the same number. By inspection the only such pairs are (2)(6) = (3)(4) = 12. Without loss of generality let ac = 2, bd = 6, ad = 3, bc = 4 and hence cd = 5. Then the remaining product is given by ab = abcd/(cd) = 12/5. The four numbers are a = sqrt((ac)(ad)/(cd)) = sqrt(6/5), b = sqrt((bc)(bd)/(cd)) = sqrt(24/5), c = sqrt((ac)(cd)/(ad)) = sqrt(10/3) and d = sqrt((bd)(cd)/(bc)) = sqrt(15/2).

3. #### Javier Campos:

Four positive numbers in Naturals mod 7.
The last product is 1, and the original four numbers are 1,2,3,4, all of them in Nat mod 7.

4. #### Javier Campos:

In previous solution, consider 0 included in Naturals 🙂

5. #### Oscar Cunningham:

Looks like I unintentionally lost some generality.