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Using prime factorization, find the HCF and LCM of: 21, 28, 36, 45

seo-qna
Last updated date: 21st Jul 2024
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Answer
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Hint:
Here, we will first find the factors of all the numbers separately by the method of prime factorization. Then we will list all the factors of the given numbers. Using the concept HCF and LCM we will find the required values. Prime Factorization is a method of finding the factors of the given numbers in terms of prime numbers.

Complete step by step solution:
We are given with the numbers 21, 28, 36, 45.
Now, we will find the factors of all the numbers using the method of prime factorization
Now, we will find the factors of 21 using prime factorization.
 We can see that 21 is an odd number, so dividing it by the least odd prime number 3, we get
\[21 \div 3 = 7\]
Now as we have obtained our quotient as prime number, we will not factorize it further.
 Thus, the factors of 21 are 3 and 7 and can be expressed as:
\[21 = 3 \times 7\]
Now, we are finding the factors of 28 using prime factorization. Therefore, we get
 We can see that 28 is an even number, so dividing it by the least prime number 2, we get
\[28 \div 2 = 14\]
Now we will divide 14 by 2. Therefore, we get
\[14 \div 2 = 7\]
Now as we have obtained our quotient as prime number, we will not factorize it further.
Thus, the factors of 28 are 2,2 and 7 and can be expressed as:
\[28 = 2 \times 2 \times 7\]
Now, we will find the factors of 36.
 We can see that 36 is an even number, so dividing it by the least prime number 2, we get
\[36 \div 2 = 18\]
Now we will divide 14 by 2. Therefore, we get
\[18 \div 2 = 9\]
Dividing 9 by the next least prime number 3, we get
\[9 \div 3 = 3\]
Now as we have obtained our quotient as prime number, we will not factorize it further.
Thus the factors of 36 are 2,2,3 and 3 and can be expressed as:
\[36 = 2 \times 2 \times 3 \times 3\]
Now, we will find the factors of 45 using prime factorization.
 We can see that 45 is an odd number, so dividing it by the least odd prime number 3, we get
\[45 \div 3 = 15\]
Now we will divide 15 by 3. Therefore, we get
\[15 \div 3 = 5\]
Now as we have obtained our quotient as prime number, we will not factorize it further.
Thus the factors of 36 are 3,3 and 5 and can be expressed as:
\[45 = 3 \times 3 \times 5\]
Thus the factors of all the numbers are represented with the same bases as:
\[\begin{array}{l}21 = {2^0} \times {3^1} \times {5^0} \times {7^1}\\28 = {2^2} \times {3^0} \times {5^0} \times {7^1}\\36 = {2^2} \times {3^2} \times {5^0} \times {7^0}\\45 = {2^0} \times {3^2} \times {5^1} \times {7^0}\end{array}\]
Now, we will find the HCF for the given numbers from the factors.
Highest common factor is a factor which is common for all the factors.
Thus, we get
HCF of \[\left( {21,28,36,45} \right) = {2^0} \times {3^0} \times {5^0} \times {7^0}\]
We know that any number raised to the power zero is one. So, we get
\[ \Rightarrow \] HCF of \[\left( {21,28,36,45} \right) = 1\]
Now, we will find the LCM for the given numbers from the factors.
Least Common Multiple is a multiple which is divisible by all the numbers.
Thus, we get
LCM of \[\left( {21,28,36,45} \right) = {2^2} \times {3^2} \times {5^1} \times {7^1}\]
Applying the exponent on the terms, we get
\[ \Rightarrow \] LCM of \[\left( {21,28,36,45} \right) = 4 \times 9 \times 5 \times 7\]
Multiplying the terms, we get
\[ \Rightarrow \] LCM of \[\left( {21,28,36,45} \right) = 1260\]

Therefore, the HCF of \[\left( {21,28,36,45} \right)\] is 1 and the LCM of \[\left( {21,28,36,45} \right)\] is 1260.

Note:
The Highest Common Factor (H.C.F) of two numbers is defined as the greatest number which divides exactly both the numbers. The Least Common Multiple (L.C.M) of two numbers is defined as the smallest number which is divisible by both the numbers. HCF can be found by multiplying the factors with the least exponent common for all the factors and LCM can be found by multiplying the factors with the highest exponent from all the factors.