Question

If A,B and C are three numbers, such that the L.C.M. of A and B is B and the L.C.M. of B and C is C , then the, L.C.M of A,B and C is
(a) A
(b) B
(c) C
(d) \[\dfrac{A+B+C}{3}\]

Answer 1

Hint: We have L.C.M of the two numbers A and B as B and L.C.M of B and C is C. We know that if the LCM of the given two numbers is equal to one of the numbers then that number is multiple of the other number. Here, we have L.C.M of A and B as B. So, B is multiple of A.
L.C.M of B and C is C. So, C is also a multiple of B. The L.C.M of three numbers where the second and third number are the multiple of the first number is the highest multiple of the first number. Now, solve further and find the value of L.C.M of the three numbers A,B and C.

Complete step-by-step answer:

According to the question, it is given that the L.C.M of the given two numbers A and B is B. Also, the L.C.M of the numbers A and C is C.
We know that if the LCM of the given two numbers is equal to one of the numbers then that number is multiple of the other number. So, we can say that B is multiple of A.
Let us assume that x times of A is equal to B.
\[\operatorname{B}=xA\] …………….(1)
We also have the L.C.M of B and C as C. So, C is also a multiple of B.
Let us assume that y times of B is equal to C.
\[C=yB\] ………………(2)
Using equation (1) and equation (2), we can write the number C as
\[\begin{align}
  & C=yB \\
 & \Rightarrow C=y(xA) \\
\end{align}\]
\[\Rightarrow C=xyA\] …………………(3)
Now, we have to find the L.C.M of the A,B and C.
In other words, we can say that we have to find the L.C.M of the number A, xA, and xyA.
We can see that all numbers are multiple of A. The L.C.M of three numbers where the second and third number are the multiple of the first number is the highest multiple of the first number. Out of three numbers A, xA, and xyA, we have xyA as the highest multiple of the first number which is A.
So, xyA is the L.C.M of A, xA, and xyA.
Using equation (2), we can write xyA as C.
From equation (1), we have \[\operatorname{B}=xA\] .
Therefore, L.C.M of A,B and C is C.
Hence, option C is correct.

Note: In this question, one may write A as \[A=xB\] and B as \[B=yC\] , which is wrong. It means that A is a multiple of B and C is multiple of B, which is completely wrong. L.C.M of A and B is B. So, B is multiple of A and we can write B as \[\operatorname{B}=xA\] . Similarly, C can be written as \[C=yB\] .