# $\mathop{\text{lcm}}[n,100] = \gcd(n,100)+450~?$

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.

• try $100n=gcd(100,n)^2+450gcd(100,n)$ Nov 3, 2017 at 12:21

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 $2$

• I'm able to see that n=250 is a solution, but I'm not sure if I can find any other solutions. Nov 3, 2017 at 12:34
• @ddswsd, If $(m,2)=1;m$ is odd lcm$(m,2)=9+1\implies m|10\implies m=5$ Now if $(m,2)=1;m$ is even lcm$(m,2)=9+2$ which is odd, but lcm$(m,2)$ is even as $m$ is even Nov 3, 2017 at 12:37

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 $n=250$.

$\text{lcm}(n,100)=L;\;\text{gcd}(n,100)=G$

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

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.