Question

Boris choos three diffrent numbers.the sum of the three numbers is 36.One of trh numbers is a cube number.the other two number are factors of 20.

Find the three numbers choosen by boris.write the numbers in ascending order.

Answer 1

4,5,27

Problem:

Boris chose three different numbers.

The sum of the three numbers is 36.

One of the numbers is a perfect cube.

The other two numbers are factors of 20.

Explanation:

Let's pretend those numbers are:

`a,b, \text{ and } c`.

We are given the sum is 36:
`a+b+c=36`.

One of our numbers is a perfect cube.
`a=n^3` where
`n` is an integer.

The other two numbers are factors of 20.
`bk=20` and
`ci=20` where
`a,c,i, \text{ and } k \text{ are integers}`.

`n^3+(20)/(k)+(20)/(i)=36`

From here I would just try to find numbers that satisfy the conditions using trial and error.

`3^3+(20)/(2)+(20)/(2)`

`27+10+10`

`47`

`3^3+(20)/(4)+(20)/(5)`

`27+5+4`

`36`

So I have found a triple that works:

`27,5,4`

The numbers in ascending order is:

`4,5,27`