## What are the Numbers?

Another cute puzzle found on Facebook.

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Puzzle.A teacher wrote four positive numbers on the board and invited his students to calculate the product of any two. The students calculated only five of six products and these are the results: 2, 3, 4, 5, 6. What is the last product? What are the original four numbers?

## 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)}

13 October 2019, 5:04 am## 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).

13 October 2019, 5:29 am## Javier Campos:

Four positive numbers in Naturals mod 7.

13 October 2019, 6:37 amThe last product is 1, and the original four numbers are 1,2,3,4, all of them in Nat mod 7.

## Javier Campos:

In previous solution, consider 0 included in Naturals 🙂

13 October 2019, 6:45 am## Oscar Cunningham:

Looks like I unintentionally lost some generality.

13 October 2019, 5:57 pm## Oscar:

What is the right answer?

18 December 2019, 11:28 am## Cheburashka:

If you want solutions to be rational, you may ask instead that 4, 9, 16, 25, 36 were the numbers on the board.

7 August 2020, 7:02 am