Answer to Question #107836 in Algebra for Sk

Question

We have 4 positive numbers whose paired products are 2, 3, 4, 5, 6 and x. Find the 4 numbers and x.

Answer 1

Let's denote the four numbers we are looking for by a, b, c and d respectively.

We do not know which of the pairs of numbers give the corresponding products. Therefore only the general formulas are valid, in which we denote the products by unknown p₁, p₂, p₃, p₄, p₅ and X (the equations are enumerated for convenience)

ab=p₁ (1),

ac=p₂ (2),

ad=p₃ (3),

bc=p₄ (4),

bd=p₅ (5),

cd=X (6),

where we can assign the expression for the product X arbitrarily.

From (1) we get b=p₁/a (7),

from (2) we get c=p₂/a (8),

from (13) we get d=p₃/a (9).

From (4), (7) and (8) we get a²=p₁p₂/p₄ (10),

from (5), (7) and (9) we get a²=p₁p₃/p₅ (11),

from (6), (8) and (9) we get X=p₂p₃/a² (12).

From (10) and (11) we can see that p₂/p₄=p₃/p₅ (13)_.

If we take into account the numerical values of the products (2, 3, 4, 5, 6), then the equality (13) is possible only in two ways:

Case 1: p₂=2, p₄=3, p₃=4 and p₅=6

Case 2: p₂=2, p₄=4, p₃=3 and p₅=6

Other permutations are equivalent.

Consider case 1: p₂=2, p₄=3, p₃=4 and p₅=6. In this case p₁=5.

From (10) we get a²=p₁p₂/p₄,

a²=5*2/3.

a=$\sqrt{\frac{10}{3}}$ because according to the condition of the problem the numbers are positive.

From (7) we get b=p₁/a=5/$\sqrt{\frac{10}{3}}$ =5$\sqrt{\frac{3}{10}}$ =$\sqrt{\frac{15}{2}}$ ,

from (8) we get c=p₂/a=2/$\sqrt{\frac{10}{3}}$ =2$\sqrt{\frac{3}{10}}$ =$\sqrt{\frac{6}{5}}$ ,

from (9) we get d=p₃/a=4/$\sqrt{\frac{10}{3}}$ =4$\sqrt{\frac{3}{10}}$ =$\sqrt{\frac{24}{5}}$ .

And from (12) we get X=p₂p₃/a²=2*4/(10/3)=12/5.

Consider case 2: p₂=2, p₄=4, p₃=3 and p₅=6.

In this case p₁=5 again.

From (10) we get a²=p₁p₂/p₄,

a²=5*2/4,

a=$\sqrt{\frac{5}{2}}$ because according to the condition of the task the numbers are positive

From (7) we get b=p₁/a=5/$\sqrt{\frac{5}{2}}$ =5$\sqrt{\frac{2}{5}}$ =$\sqrt{10}$ ,

from (8) we get c=p₂/a=2/$\sqrt{\frac{5}{2}}$ =2$\sqrt{\frac{2}{5}}$ =$\sqrt{\frac{8}{5}}$ ,

from (9) we get d=p₃/a=3/$\sqrt{\frac{5}{2}}$ =3$\sqrt{\frac{2}{5}}$ =$\sqrt{\frac{18}{5}}$ .

And from (12) we get X=p₂p₃/a²=3*6/(15/2)=12/5 again.

So, we have two solutions of this problem.

Answer:

1: $\sqrt{\frac{10}{3}},$ $\sqrt{\frac{15}{2}},$ $\sqrt{\frac{6}{5}},$ $\sqrt{\frac{24}{5}}$ and the product is 12/5.

2: $\sqrt{\frac{5}{2}},$ $\sqrt{10},$ $\sqrt{\frac{8}{5}},$ $\sqrt{\frac{18}{5}}$ and the product is 12/5.