Find the LCM of 3 numbers given HCF of each 2. 
Answer is 
I was totally confused when I saw the question. I never encountered a question like this.
Can anyone tell me the way to solve this.
I tried every method I could find but hard luck.
 A: Hint: Proceed one prime at a time. Let's deal with $2$.
The highest power of $2$ that divides $a$ and $b$ is $2^3$.
But the highest power of $2$ that divides $b$ is at least $2^4$, since $\gcd(b,c)$ is divisible by $2^4$.
So the highest power of $2$ that divides $a$ is $2^3$.
Since $2^4$ divides both $b$ and $c$, it follows that $abc$ is divisible by $2^{3+4+4}=2^{11}$.
But the highest power of $2$ that divides $abc$ is $2^{11}$.
So the highest power of $2$ that divides $b$ is $2^4$, and the same is true of $c$.
So the powers are $2^3$, $2^4$, $2^4$, which means that the exponent of $2$ in the LCM is $\max(3,4,4)$, which is $4$.
Continue, reasoning your way with $3$, with $5$, with $7$.
A: As Andre Nicolas says , proceed one prime at a time.
Lets take $3$.    
The highest power of $3$ that divides $a$ and $b$ & $a$ and $c$ & $b$ and $c$ is $3^2$.
So at least $3^6$ should divide $abc$.
But it is given that $3^{10}$ divides $abc$.
So let's assume , without the loss of generality , that $3^6$ divides $b$.
The $HCF$ of $a$ and $b$ & $b$ and $c$ is still $3^2$ .  
Similarly $5^4$ divides $a$, $5^5$ divides $c$ and $5^8$ divides $b$. $( 4+5+8=17)$  
Also $7^1$ divides both $a$ and $b$. $( 1+1=2)$ 
Then we get the $LCM$ as $2^43^65^87^1$.
