Find all positive integers $a$ such that: $lcm(120,a) = 360$ and $gcd(450,a)=90$

Question

Find all positive integers $a$ such that: $lcm(120,a) = 360$ and $gcd(450,a)=90$

I started by factoring
$$450 = 2\times 3\times3\times5\times5$$
and
$$90 = 2 \times3\times3\times5$$

Since $gcd(450,a)=90$ than

$$a = 90m$$ where $m$ is not a multiple of $5$

On the other side:
$$lcm(120,a) = 360$$

Which means:

$$360 = \frac{120\cdot a}{gcd(120,a)}$$

Substituting $a$

$$360 = \frac{120\cdot 90m}{gcd(120,90m)}$$

$$gcd(120,90m) = \frac{120\cdot 90m}{360}$$

$$gcd(4,3m) = \frac{120\cdot 90m}{360\cdot 30}$$

$$gcd(4,3m) = m$$

So $$m|4$$

Which implies that $m \in \{1,2,4\}$

Finally $a \in \{90,180,360\}$

I know the result is right, because my book says so, but I only arrived at the solution by working backwards from the answer in the book. And my resolution does not seems very mathematical (I think I assumed things that I could not assume. I would be thankful if you could also point my mistakes). Can someone me a better way to solve this problem? My book has lots of problems involving variables and gcd and lcm. I am looking for a method or at least a logic to solve this kind of problem.

Answer 1

Since your answer is correct, I would assume the overall steps are also correct, but I can also look more closely if you want :)

Here's what I would've done to arrive at the solution:

• Firstly, think about definition of $\gcd$. It is the greatest common divisor, so $\gcd(450, a) = 90 \implies 90$ is a divisor of $a$ i.e. $90 \mid a$.
• Similarly, $\text{lcm}(120, a) = 360$ implies that $a$ is divisor of $360$ i.e. $a \mid 360$.

These two constraints are strong enough to narrow down our candidates to $a \in \{90, 180, 360\}$. Now it suffices to check that they work, which you can do yourself :P