If $\gcd(x,y)=1379,\operatorname{lcm}(x,y)=n! $ then find $n$ ($x,y,n$ are positive integers).
I tried using the relation: $\gcd(x,y)\cdot\operatorname{lcm}(x,y)=xy$ and also wrote $1379=7\cdot197$ but nothing more.

  • 1
    $\begingroup$ Since $197$ is prime you need $n\ge197$. Can you see how to choose $x,y$ so that $n=197$ works? $\endgroup$
    – almagest
    Jun 14 '16 at 6:14
  • $\begingroup$ @almagest Why is it necessary that: $n>=197$? $\endgroup$ Jun 14 '16 at 6:16
  • $\begingroup$ Because none of $1,2,\dots,n$ can be divisible by a prime greater than $n$. $\endgroup$
    – almagest
    Jun 14 '16 at 6:22
  • $\begingroup$ @almagest you mean this equation $7*197*n!=xy$ implies $n>=197$??!! I think it's not necessary that $n$ be divisible by $197$,on the other hand $x$ or $y$ should be... $\endgroup$ Jun 14 '16 at 6:25
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    $\begingroup$ The gcd is divisible by 197, so the lcm is divisible by 197, so $n!$ is divisible by 197. Since 197 is prime that means that at least one of $1,2,\dots,n$ is divisible by 197. That means that $n\ge197$. $\endgroup$
    – almagest
    Jun 14 '16 at 6:28

Since $\gcd(x,y)=7\cdot 197$, we can assume $x=7\cdot 197a$, and $y=7\cdot 197b$, where $\gcd(a,b)=1$. Then $\text{lcm}(x,y)=7\cdot 197\cdot ab=n!$. Thus, $n\geqslant 197$. Actually, we can assign different prime factors of $\frac{n!}{7\cdot 197}$ to $a$ and $b$ respectively to make sure they are coprime.


$n$ can be any integer which is no smaller than $197$.


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