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$A$ and $B$ are positive integers such that $$\gcd(A,B)=9, \text{lcm}(A,B)=180, A<B.$$ How many such $A$'s are there?

So from $\text{lcm}(A,B)=\frac{AB}{\gcd(A,B)} \Rightarrow 180 = \frac{AB}{9} \Rightarrow 1620=AB.$

Prime factoring $1620= 2^2\cdot3^4 \cdot5 = 4\cdot81\cdot5$, from here it seems that $81$ is the largest prime factor and the two other possbile prime factors are $4$ and $5$. Just following my intuition this got me to the right answer which was $2$, however I cannot find any justification for this. Is my reasoning correct or could the be a setting where this would fail? If so what would be the alternatives?

Air Mike
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2 Answers2

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WLOG

$\dfrac Aa=\dfrac Bb=9\implies(a,b)=1$

$180=[A,B]=[9a,9b]=9[a,b]\iff[a,b]=20$

Now as $a<b$ and $(a,b)=1,$ so $a\in\{1,4\}$

lab bhattacharjee
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Consider the prime factors of $1620$ being distributed into $A$ and $B$

Since 9 is the gcd, each of $A$ and $B$ have to have $3^2$ as part of their prime factorisation, and nothing else common

$$A= 3^2 X$$ $$B = 3^2Y$$

Now what is remaining? $2^2 , 5$ - since both twos have to go to either A or B - you can have only two possible answers for $A$ such that $A<B$ :

$$A = 3^2\cdot 2^2 \space\text{or}\space 3^2$$ $$B = 3^2\cdot 5 \space \text{or} \space3^2\cdot 2^2\cdot 5$$

Dhanvi Sreenivasan
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