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:
[tex]a,b, \text{ and } c[/tex].
We are given the sum is 36: [tex]a+b+c=36[/tex].
One of our numbers is a perfect cube. [tex]a=n^3[/tex] where [tex]n[/tex] is an integer.
The other two numbers are factors of 20. [tex]bk=20[/tex] and [tex]ci=20[/tex] where [tex]a,c,i, \text{ and } k \text{ are integers}[/tex].
[tex]n^3+\frac{20}{k}+\frac{20}{i}=36[/tex]
From here I would just try to find numbers that satisfy the conditions using trial and error.
[tex]3^3+\frac{20}{2}+\frac{20}{2}[/tex]
[tex]27+10+10[/tex]
[tex]47[/tex]
[tex]3^3+\frac{20}{4}+\frac{20}{5}[/tex]
[tex]27+5+4[/tex]
[tex]36[/tex]
So I have found a triple that works:
[tex]27,5,4[/tex]
The numbers in ascending order is:
[tex]4,5,27[/tex]
