Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term?

Question

Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Please explain.

I found $n=2$ be a solution. Is it correct?

Answer 1

Your example $n=2$ is correct. We will find all possible examples.

Suppose that $d$ divides $9n+5$ and $10n+3$. Then $d$ divides $$10(9n+5)-9(10n+3),$$ so $d$ divides $23$.

Since $23$ is prime, the only conceivable (positive) common divisors of $9n+5$ and $10n+3$ are $1$ and $23$. We will find the values of $n$ for which $23$ is a common divisor.

To make sure that $23$ divides both, all we need is to make sure that $23$ divides $9n+5$. This is because from $10(9n+5)-9(10n+3)=23$ we can conclude that if $23$ divides $9n+5$, it must divide $10n+3$.

You noticed that $23$ divides $9n+5$ when $n=2$. It follows that $23$ divides $9n+5$ if and only if $n=2+23k$ for some integer $k$. For if $23$ divides $9n+5$, then $23$ divides $9n+5-23$, so $23$ divides $9(n-2)$, and therefore $n-2$ is a multiple of $23$. Conversely, if $n-2$ is a multiple of $23$, say $n-2=23k$, then a calculation shows that $23$ divides $9n+5$.