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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.

Bill Dubuque
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user3347814
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    @zwim False. For $m = 2$, $\text{lcm}(120, 180) = 360 \neq 720$. – Gareth Ma May 27 '22 at 15:42
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    One simple way that always works for problems of this type is to write $,a = 2^i 3^j 5^k,$ then find constraints on $,i,j,k,$ using the formulas for gcd, lcm in terms of prime factorizations. We have many prior worked problems of this sort already. Did you search? – Bill Dubuque May 27 '22 at 15:47

3 Answers3

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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

Gareth Ma
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Your answer is correct. The fact of $gcd(a,450)$ implies that exists a integer $d$ such that $d|a$ and d|450 and $d$ is minimum. In this case this value $d=90$, as 90 divides $a$, $a$ must be a multiple of 90, hence, $a=90m$, where $m$ is an integer. Using this and replacing on $lcm(120,a)$, we get that $lcm(120,90m)$. By definition, $lcm(120,90m)=\displaystyle\frac{|120(90m)|}{gcd(120,90m)}=360$. Rearranging $gcd(120,90m)=\displaystyle\frac{120(90)|m|}{360}=30|m|$.

By properties of $\gcd$, write $\gcd(120,90m)=30|m|$ is the same that $(4,3m)=|m|.$ This implies that $|m||3m$ (obviously) and $|m||4$, this last equality implies that $m=-4,-2,-1,1,2,4$. Finally, $a=-360,180,-90,90,180,360$.

amWhy
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Omar Iñaki Carbajal Bonal
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Here’s an alternative method that generalizes well to any such problem.

You can first see what primes will be relevant here. By factoring all of the numbers, you can see that we care about 2,3, and 5.

Look at each equation for each prime.

$120$ And $360$ have 3 factors of $2$, so since $lcm(120,a)=360$, that means that $a$ has at most three prime factors of 2 Looking at the other formula gives $a$ has at least one factor of 2.

Similarly going through for 3 gives that the maximum of the number of 3’s in 120 and $a$ is the number in $360$. Since $120$ has exactly one and 360 has two, this means that $a$ has two.

Similarly, the gcd formula gives that $a$ has exactly one 5.

No other primes are in $a$ by looking at the lcm formula.

Therefore, $a$ must be 2,4, or 8 times 9 times 5 giving 90,180, or 360 as desired.

Eric
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