Is this a trick question?

"Which pairs of integers $a$ and $b$ have greatest common divisor $18$ and least common multiple $540$?"

There is a theorem that states that $[a,b](a,b)=ab$, where $[a,b]$ stands for the least common multiple of $a$ and $b$, and $(a,b)$ stands for the greatest common divisor of $a$ and $b$.

So, it follows that$$[a,b](a,b)=ab,$$$$540\cdot18=a\cdot b.$$In other words, $a=540$ and $b=18$ would suffice.

Am I missing something?

Question

Is this a trick question?

"Which pairs of integers $a$ and $b$ have greatest common divisor $18$ and least common multiple $540$?"

There is a theorem that states that $[a,b](a,b)=ab$, where $[a,b]$ stands for the least common multiple of $a$ and $b$, and  $(a,b)$ stands for the greatest common divisor of $a$ and $b$.

So, it follows that$$[a,b](a,b)=ab,$$$$540\cdot18=a\cdot b.$$In other words, $a=540$ and $b=18$ would suffice.

Am I missing something?

jagatuba · Answer

I don't think you are missing anything. The GCD of 18=18 and the LCM of 540=540, so as far as the problem goes with the figures you are given the theorem holds true.

anonymous · Answer

Thank you.

mathmate · Answer

It is probably a trick question, because it asked
"Which pairS of integers a and b ..."
implies that you need to find ALL the pairs.
Your logic is perfectly accurate, so let's continue from there.

Break 540 into factors
$ 540=(3^2.2).3.2.5 = 18.3.2.5$
So the pairs of A and B could be taken from the set $S=\{18,1,3,2,5\} $ such that
$A \cup B = S \ and\  A \cap B = \{18\}$
From this, we can find pairs of a,b such as
(18,540), (36,270),(54,180),(90,108), all satisfying the above conditions.