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

Answer:

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.

Step-by-step 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 $a,c,i, \text{ and } k \text{ are integers}$.

$n^3+\frac{20}{k}+\frac{20}{i}=36$

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

$3^3+\frac{20}{2}+\frac{20}{2}$

$27+10+10$

$47$

$3^3+\frac{20}{4}+\frac{20}{5}$

$27+5+4$

$36$

So I have found a triple that works:

$27,5,4$

The numbers in ascending order is:

$4,5,27$