If $a=7!$, $b=_{13}P_k$, $\frac{ab}{\operatorname{lcm}(a,b)}=120$, find the $k$

Question

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?

Answer 1

This was based on the original post

The problem as stated is incorrect or there is a typo. If $\frac{ab}{\gcd(a,b)}=120$, then that would mean $\text{lcm}(a,b)=120$. But $120=\text{lcm}(a,b) \geq a=7!$. This is not possible.

After the original post got modified:

Perhaps, the $\gcd(a,b)=120$. In which case we can do the following: \begin{align*} \gcd(a,b)&=120\\ \gcd(7!,b)&=120\\ \gcd\left(42,\frac{b}{120}\right)&=1 \end{align*} This means $\frac{b}{120}$ is (at least) not divisible by $2,3$ and $7$, where $b=\frac{13!}{(13-k)!}$. We know that $13!=2^{10} \cdot 5^2 \cdot 7^{1} \dotsb$. So $$\frac{b}{120}=\frac{13!}{(2^3\cdot 3 \cdot 5) \, (13-k)!}=\frac{2^{7} \cdot 3^4\cdot 5^1 \cdot 7^{1} \dotsb}{(13-k)!}$$

This means our $k$ should be such that the prime factorization of $(13-k)!$ should also have exactly these powers of the primes $2,3$ and $7$.

Since $2$ should only appear with exponent $7$ in the prime factorization of $(13-k)!$, so $13-k \geq 8 \implies k \leq 5$ and $13-k \leq 9 \implies k \geq 4$.

Since $3$ should only appear with exponent $4$ in the prime factorization of $(13-k)!$, so $13-k \geq 9 \implies k \leq 4$.

Thus $\color{red}{k=4}$ will satisfy the gcd condition.

Answer 2

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.