Find number of possible triplets $(a,b,c)$ if
$LCM\left(a,b\right)=5^2\times 3^4$
$LCM\left(b,c\right)=5^3\times 3^4$
$LCM\left(a,c\right)=5^3\times 3^4$
My Attempt
Since $LCM\left(a,b\right)=5^2\times 3^4$. Power of $5$ in both $a$ and $b$ therefore can never be $3$. So, power of $5$ in $c$ must be $3$.
But after this I am not able to solve
I'm a 13 year old who loves maths, so i might be wrong, but First, just like you noticed, the p-adic valuation for c is $\nu_{5}=3$, now, we can notice that between a and b, one of them must be $\nu_{5}=2$, and the other is free to be $\nu_{5}\in\{0,1,2\}$, which gives us $3×2=6$ ways of distributing the power of 5 Now for the power of 3, we take a and b, since their LCM has $3^4$, at least one of them has $\nu_{3}=4$. If it's a, then b has $\nu_{3} \in\{0,1,2,3,4\}$. Then if b has $\nu_{3}=4$, then c has $\nu_{3}\in\{0,1,2,3,4\}$, but if b has $\nu_{3}\in\{0,1,2,3\}$, then c must have $\nu_{3}=4$ thus at least two numbers have $\nu_{3}=4$, and the third one has $\nu_{3}\in\{0,1,2,3,4\}$ thus there are $3×5=15$ ways to rearrange here. From all of this, we get that there are $6×15=75$ triplets that satisfy this system of equations.