Question

The HCF and the LCM of 12, 21, 15 respectively are
A) 3, 140
B) 12, 420
C) 3, 420
D) 420, 3

Answer 1

Hint:
Here, we are required to find the H.C.F. and the L.C.M. of the given three numbers. We will find the factors of the three numbers by using the prime factorization method. Then multiplying the factors that are common in all the three numbers will give us the highest common factor (H.C.F.) of the three numbers. Now multiplying all the factors of the three numbers by taking the highest power among the common factors, we will be able to the required L.C.M. of the three numbers.

Complete step by step solution:
First of all, we have to find the H.C.F. of three numbers. A Highest common factor (H.C.F.) is the greatest number that is a factor of two or more numbers. This is also known as the Greatest Common Factor (G.C.F.).
By using the prime factorization method, we will find the factor of the given three numbers.
We will first find the factors of 12.
Now we can see 12 is an even number, so dividing it by the least prime number 2, we get
$$12÷2=6$$
Dividing 6 by 2, we get
$$6÷2=3$$
As we can see that the obtained quotient is a prime number, so we will not divide the number further.
Therefore, $$12=2\times 2\times 3=2^2\times 3$$
We will first find the factors of 12.
Now we can see 21 is an odd number, so dividing it by the least odd prime number 3, we get
$$21÷3=7$$
As we can see that the obtained quotient is a prime number, so we will not divide the number further.
Therefore, $$21=3\times 7$$
We will first find the factors of 12.
Now we can see 15 is an odd number, so dividing it by the least odd prime number 3, we get
$$15÷3=5$$
As we can see that the obtained quotient is a prime number, so we will not divide the number further.
Therefore, $$15=3\times 5$$
As we can see, in all the three numbers, we can find the number 3 multiplying by other factors. Hence, 3 is the highest common factor of the given three numbers.
The prime factors of the three numbers are
$$12=2^2\times 3$$
$$21=3\times 7$$
$$15=3\times 5$$
Hence, L.C.M. of these three numbers $$=2^2\times 3\times 5\times 7$$
This is because in L.C.M. we take all the factors present in three numbers and the highest power of the common factors respectively.
Thus, L.C.M. of the given 3 numbers $$=21\times 20=420$$
Therefore, the required H.C.F. and L.C.M. of 12, 21 and 15 are 3, 420 respectively.

Hence, option C is the correct answer.

Note:
In this question, we are required to express the given numbers as a product of their prime factors in order to find their HCF and LCM respectively. Hence, the prime factors are those factors which are greater than 1 and have only two factors, i.e. factor 1 and the prime number itself.
Now, in order to express the given numbers as a product of their prime factors, we are required to do the prime factorization of the given number. Now, factorization is a method of writing an original number as the product of its various factors. Hence, prime factorization is a method in which we write the original number as the product of various prime numbers.