$\mathop{\text{lcm}}[n,100] = \gcd(n,100)+450~?$ What is the sum of all positive integers $n$ that satisfy $$\mathop{\text{lcm}}[n,100] = \gcd(n,100)+450~?$$
This problem seems really interesting, any hints are greatly appreciated. I did prime factorization of 100 to try and gain some information.
 A: Hint: Write $d=\gcd(n,100)$. Then $n=dn'$ and $\operatorname{lcm}(n,100)=100n'$. Therefore, $100n' = d + 450$ and so $d$ is a multiple of $\gcd(100,450)$.
Solution:

 $d$ is a multiple of $50$ and also a divisor of $100$. Hence, $d=50$ or $d=100$. But $d$ cannot be $100$ because $450$ is a not a multiple of $100$. Thus, $d=50$ and $n'=5$, which gives $n=250$. 

A: Hint:
gcd$(n,100)=$lcm$(n,100)-450$
If $p=$lcm$(n,100),$
As $100|p,$ gcd$(p,450)$ must divide $(100,450)=50$
$\implies50$ must divide $n,$ let $n=50m$
$50$gcd$(m,2)=50$lcm$(m,2)-450$
$\iff$gcd$(m,2)=$lcm$(m,2)-9$
Now gcd$(m,2)$ should be $1$ or $2$
A: $\text{lcm}(n,100)=L;\;\text{gcd}(n,100)=G$
product of $lcm$ by $gcd$ is equal to the product of the two numbers
$LG=100n;\;L=G+450$
$G(G+450)=100n$
$G^2+450G-100n=0$
$G=-225+\sqrt{225^2+100n}$
$n=250;\;G=50;\;L=500$
No other solutions because the $GCD$ must be less than $100$
Hope this is useful
A: Hint: Suppose $n>100$. Multiply both side of the equation by $gcd(n,100)$ to get:
$$
100n = gcd(n,100)^2 +450 gcd(n,100) < 100^2 +100\cdot 450 \implies n<550
$$

 Now, since we have $50 |450-lcm(n,100)=gcd(n,100)$, We can deduce that $50|n$. Since $lcm(n,100)>450$ we can rule out $50,100,150,200,300,400$ and  we are left with the candidates $250,350,450,500$. Checking all the numbers yields that the only possible answer is 250.

