Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.

Question

Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.

Attempt:

$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$

$$n=1: (4-5,45)=1\quad \checkmark\\ n=2: (3,45)=3\quad \times\\ n=3: (7,45)=1\quad \checkmark\\ n=4: (11,45)=1\quad \checkmark\\ n=5: (15,45)=15\quad \times\\ n=6: (19,45)=1\quad \checkmark\\ n=7: (23,45)=1\quad \checkmark\\ n=8: (27,45)=9\quad \times\\ \vdots$$ So the answer is that it can't be reduced for $n=1,3,4,6,7,..$ i.e

$$\bigg\{n\bigg|n\notin \begin{cases}a_1=2\\a_n=a_{n-1}+3\end{cases}\bigg\}$$

I want to verify that my solution is correct

Answer 1

We have that $\gcd(pq + r,p) = \gcd(p,r)$ and since $12n - 60 = 3(4n-5)+45$, we have that $\frac{4n-5}{60-12n}$ can't be reduced if and only if $\gcd(4n-5,45) = 1$ and since $45=3^2\cdot 5$, this happens if and only if $3\nmid 4n-5$ and $5\nmid 4n-5$.

Now,

$4n-5\equiv 0 \pmod 3\iff 4n \equiv 8 \pmod 3 \iff n\equiv 2\pmod 3$ (last equivalence is due to $\bar 4$ being invertible in $\mathbb Z/3\mathbb Z$)

$4n-5\equiv 0\pmod 5\iff 4n\equiv 0\pmod 5 \iff n\equiv 0\pmod 5$ (last equivalence is due to $\bar 4$ being invertible in $\mathbb Z/5\mathbb Z$)

Thus, $\frac{4n-5}{60-12n}$ can't be reduced if and only if $n\not\equiv 2\pmod 3$ and $n\not\equiv 0\pmod 5$.

Answer 2

If $d$ is the common divisor of $4n-5$ and $60-12n$, then it should also divide $3(4n-5)+(60-12n) =45$. Thus when numerator and denominator are not divisible by 3 and 5, the fraction is irreducible. Hence $n \not \equiv 2 \pmod 3$ and $n \not \equiv 0 \pmod 5$.

Answer 3

I want to verify that my solution is correct

Your solution is not correct.

For $m\not=1$, $$\frac{4\times\color{red}{5m}-5}{60-12\times\color{red}{5m}}=\frac{5(4m-1)}{5(12-12m)}$$ can be reduced.

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If $n$ is of the form $3k+2$, then $$\frac{4(3k+2)-5}{60-12(3k+2)}=\frac{3(4k+1)}{3(-12k+12)}$$ can be reduced.

If $n$ is not of the form $3k+2$, then since $4n-5$ is not divisible by $3$, we get $$\gcd(4n-5,12(5-n))=\gcd(4n-5,4(5-n))=\gcd (4n-5,15)=\gcd(4n-5,5)$$

Now note that $$\gcd(4n-5,5)=1\iff n\not\equiv 0\pmod 5$$ Hence, the answer is that $$n\not\equiv 2\pmod 3\quad\text{and}\quad n\not\equiv 0\pmod 5.$$

Answer 4

It apears that you attempted to check when $\,(N,45) := (4n-5,\,45)=1\,$ by a brute-force case analysis on $\,n.\,$ But the result is not correct since you didn't examine enough cases (you need to check all $\,n\pmod {45},\,$ i.e. $\,45\,$ cases). Much simpler is to use the following:

$ N$ and $\, 3^2\cdot 5\,$ have a common factor iff they have a common prime factor iff $\,3\mid N\,$ or $\,5\mid N$

because, by uniqueness of prime factorizations, the prime factors of $\,3^2\cdot 5\,$ are $\,\{3, 5\}$.