<s>[INST] If $a=7!$, $b=_{13}P_k$, $\frac{ab}{\operatorname{lcm}(a,b)}=120$, find the $k$ 
Question:
$a=7!$, $b=_{13}P_k$, $\dfrac{ab}{\operatorname{lcm}(a,b)}=120$, then find the $k$.

My attempts:
$$\frac{ab}{\operatorname{lcm}(a,b)}=\gcd{(a,b)}=120$$
$\gcd{(7!, _{13}P_k)}=120 \Longrightarrow \begin{cases} \dfrac{13!}{(13-k)!×5!}\in \mathbb {Z^+} \\ \dfrac{13!}{(13-k)!×5!} \mod 2≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 3≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 7≠0  \end{cases} \Longrightarrow \begin{cases} 
\dfrac{13×12×\cdots (13-k+1)!}{120} \in \mathbb {Z^+} \\ 
 \dfrac{13!}{(13-k)!×5!} \mod 2≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 3≠0 \\
\dfrac{13!}{(13-k)!×5!} \mod 7≠0  \end{cases} \Longrightarrow 13-k+1=10 \Longrightarrow k=4$
Is my solution correct? Do I have any missing or unnecessary/unneeded steps?
 [/INST] I think it'd be simpler to do this:
$\frac {ab}{\operatorname{lcm}(a,b)}=\gcd(a,b)$ so we have
$\gcd(7!, b) = 120$
So $\gcd(2^4\cdot 3^2\cdot 5\cdot 7, b) = 2^3\cdot 3\cdot 5$.
So $3|b$ but $9\not \mid b$ and $8|b$ but $16\not\mid b$ and $5|b$ and $7\not \mid b$.
$b =13*(13-1)*....*(13-(k-1))= \prod_{m=13-k+1}^{13} m$. Let $L = 13-k+1$ If $L \le 9$ then $9|b$.  So $L>9$.  If $10 < L < 15$ then $5\not \mid b$ so $L\le 10$.  So $L = 10$.
So $13- k+1=10$ and $k=4$.
But we must verify that $L=10$ is acceptable namely that $3|b$ (which is does as $3|12$) that $9\not \mid b$ (which it doesn't as $12$ is the only multiple of $3$ between $10$ and $13$ inclusively) that $8|b$ but $16\not \mid b$ (as $2|10$ and $4|12$ and $10, 12$ are the only two even numbers between $10$ and $13$ that is fine) and that $5|b$ (which as $5|10$ it does) and the $7\not \mid b$ (which as there are no multiples of $7$ between $10$ and $13$ it does not.
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