Question

If $r,s,t$ are different prime numbers and $a,b,c$ are the positive integers such that $LCM$ of $a,b,c$ is $r^2{s}^4{t}^2$ and $HCF$ of $a,b,c$ is $r{s}^{2}t$, then the number of ordered triplets $(a,b,c)$ is

• A. $248$
• B. $432$
• C. $648$
• D. $931$

Answer 1

Answer: (B)

$432$

Here $L.C.M$ of $a,b,c$ is $r^2{s}^4{t}^2$

Hence combination of $r,s,t$ be

for $r^2{s}^2{t}^2$ the no of combination is $3^3$

But we have $t^5$ hence we should add this combination

So total no of combination is $27\times 2=54$

Here $H.C.F$ is $r{s}^{2}t$

no of ways for $r$ is $r=r^0,r$

no of ways for $s$ is $s=s^0,s$

no of ways for $t$ is $r=t^0,t$

Hence no of ways for $H.C.F$ is $2\times 2\times 2=8$

Total no of ways for ordered triplets is $54\times 8=432$