What Integer Values Satisfy the Equation xy^2=54 with Constraints?

[chead9]

What are the possible values of y such that xy^2=54, x is less than 10, y is less than 10, and x and y are integers? How do I go about finding this answer?

 

[MarkFL]

If both $x$ and $y$ have to be integers greater than 10, then what is the smallest value for $xy^2$?

 

[chead9]

If both $x$ and $y$ have to be integers greater than 10, then what is the smallest value for $xy^2$?

I made a mistake in the post.. it was supposed to be x and y are both less than 10

 

[MarkFL]

I made a mistake in the post.. it was supposed to be x and y are both less than 10

Ah, okay...now we're in business. :)

I think I would start out by arranging the given equation as:

\(\displaystyle x=\frac{54}{y^2}\)

Now, if $x$ is to be an integer, then $y^2$ must be a factor of 54 and at the same time a perfect square. Can you think of any such numbers?

 

[MarkFL]

We have:

\(\displaystyle x=\frac{54}{y^2}\)

And since we require:

\(\displaystyle x<10\)

this means (also gven $y<10$):

\(\displaystyle \frac{54}{y^2}<10\implies 3\sqrt{\frac{3}{5}}<y<10\)

And since $y$ must be an integer, we should write:

\(\displaystyle \left\lceil3\sqrt{\frac{3}{5}}\right\rceil\le y<10\)

\(\displaystyle 3\le y<10\)

We need a number $y^2$ which is a factor of 54 and is a perfect square...so looking at the prime factorization of 54, we find:

\(\displaystyle 54=2\cdot27=2\cdot3^3=6\cdot3^2\)

Thus, we must have:

\(\displaystyle y=3\implies x=6\)