Question

Find the L.C.M of $24,36,108$ and $192$.

Answer 1

Hint: The question gives us four numbers and we need to find the Least Common Multiple or Lowest Common Multiple (L.C.M) of these numbers. To find the Least Common Multiple, we first list out the prime factors of all the given numbers. Then we choose the prime numbers with the highest powers. That is, we take the common multiple to get to the final answer. Hence, this method is called Prime factorization.

Complete step by step answer:
The given numbers are,
$$24,36,108$$ and $$192$$
Let us consider each of them, one by one. Then, we will factorize each as multiple of prime numbers.
The prime factors for $$24$$ are
$$24 = 2 \times 2 \times 2 \times 3$$
The prime factors for $$36$$ will be
$$36 = 2 \times 2 \times 3 \times 3$$
The prime factors for $$108$$ are
$$108 = 2 \times 2 \times 3 \times 3 \times 3$$
The prime factors for $$192$$ are expressed as
$$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$$
From the above, we see that $$2$$ and $$3$$ are the two prime numbers.
Now, let us see where these numbers are present with highest powers
$$2$$ has the highest power of $$6$$ in the number $$192$$
$$3$$ has $$3$$ as the highest power, in the number $$108$$
 So, we multiply these to get the L.C.M
Hence, the L.C.M will be,
$$L.C.M = {2^6} \times {3^3}$$
On multiplication, we get
$$ \Rightarrow L.C.M = 1728$$
Hence, the L.C.M of the given numbers, $$24,36,108$$ and $$192$$ is $$1728$$.

Note:
Note that the given numbers are all even numbers. So, two will definitely be a prime factor of all the given numbers. And, we also see that the sum of the digits is divisible by three. Hence, three will also be a prime factor. Be careful while choosing the highest powers. This can also be solved by long division or common division methods.