The least common multiple of 125 and an unknown number x is 45 times their greatest common divisor. Here is what I've tried:
Let $\gcd(125,x) = G$ and $\mathrm{lcm}(125,x)=L$.
We know
$$
125 \times x = G \times L \ \text{ and } L = 45G
\implies 125x = 45G^2
\implies \frac{25}{9}x = G^2.
$$
Because $G$ is an integer, so are $x$ and $G^2$, so $9|x$.
We know $125\times x = 45G^2$. The two numbers are not relatively prime, because $G \ne 1$. So the number we are looking for should at least be divisible by $5$, and hence by $45$. By trying out multiples of $45$, I get the answer as 225.
Is there a way to eliminate the numbers closer ? Even this might be easy because 125 has 3 factors but it is long way for numbers with many factors like 360. Is any other way to solve but also shorter ?
Thanks
Best Answer
You also have the solution of $$x= 45\times 125$$
Note that in this case $L=45\times 125$ and $G=125$ thus $L=45G$
In general If $G(x,y)=d$, then $x=md$ and $y=nd$ where $G(m,n)=1.$
Using this fact may help reduce calculations in more complicated cases.