Jacoby Erickson

2022-10-11

For which natural numbers a fraction can be simplifed
I want to know how to check for which natural numbers fraction $\frac{7n+19}{7n+2}$ can be simplified. The correct answer is supposed to be $n=2k-1$
I do understand that the denominator and the numerator have to have a common divisor in order to simplify the fraction, but fail to see how this particular problem should be approached.

Alannah Yang

$\frac{7n+19}{7n+2}=1+\frac{17}{7n+2}$
Thus, in order for $\frac{7n+19}{7n+2}$ to be simplified, we have that $\frac{17}{7n+2}$ can be simplified as well.
This implies that $17$ and $7n+2$ must have a common divisor that is not $1$. Since $17$ is prime, this implies that
$7n+2\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}17\right)$
This gives us that
$n\equiv 7\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}17\right)$
Thus, the answer is that $n=17k+7$ where $k\in \mathbb{Z}$, not $2k-1$ as stated, so your answer is incorrect.

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