Cubes and square

Algebra Level 1

A positive number when squared becomes 8 times another number.

And when the first number is cubed it becomes 32 times the other number.

What is the smallest possible value of such two numbers. Give the product of the two numbers.

32 16 8 64

2 solutions

Roger Erisman
Apr 18, 2016

x^2 = 8*y

x^3 = 32*y

Eqn 2 / Eqn 1 gives x = 4 which means y = 2

Product = 4 * 2 = 8

Nice and tidy solution :)

Abhay Tiwari - 5 years, 1 month ago

I grow curious, what if both were 0. XD. Then, all the conditions are satisfied. LOL

Ashish Menon - 5 years, 1 month ago

0 is not a positive number

genis dude - 4 years, 3 months ago

@Genis Dude – True, my bad sorry.

Ashish Menon - 4 years, 3 months ago

Ha ha ha right !

Abhay Tiwari - 5 years, 1 month ago

A Former Brilliant Member
Apr 1, 2017

$x=$ $\text{first number}$

$y=$ $\text{other number}$

$\text{From the problem, we have}$

$x^2 = 8y$ $\color{#D61F06}(1)$

$x^3 = 32y$ $\color{#D61F06}(2)$

$\text{Divide (2) by (1)}$ $\implies$ $\color{#3D99F6}\large x = 4$

$\text{Solve for y by substituting 4 to x in any of the equations}$ $\implies$ $\color{#3D99F6}\large y = 2$

$\text{Their product is}$ $(2)(4) =$ $\boxed{\color{#20A900}\large 8}$