
Can you find the HCF of \[1.2\] and \[0.12\]? Justify your answer.
Answer
446.4k+ views
Hint:
Here, we will first convert the decimal numbers into fractions with the same denominator. Then we will find the factors of the numerators of both the fractions by using the Prime Factorization method. Then by using the common factors, we will find the HCF of both the decimal numbers.
Complete step by step solution:
We are given the numbers \[1.2\] and \[0.12\].
First, we will convert the numbers with equal numbers of decimal digits.
So, we get \[1.20\] and \[0.12\].
Now, we will convert both the decimal numbers in the form of fraction. So, we get
\[1.20 = \dfrac{{120}}{{100}}\]
\[0.12 = \dfrac{{12}}{{100}}\]
Since both the denominators are equal, we will find the HCF of the two numbers in the numerator 120 and 12.
We will find the factors of the numerators by using the prime factorization method.
First, we will do prime factorization of 120.
We can write 120 as a product of 2 numbers. Therefore,
\[120 = 12 \times 10\]
Breaking the numbers into product of 2 numbers, we get
\[\begin{array}{l} \Rightarrow 120 = 4 \times 3 \times 5 \times 2\\ \Rightarrow 120 = 2 \times 2 \times 2 \times 3 \times 5\end{array}\]
So, we can write 120 as \[120 = {2^3} \times {3^1} \times {5^1}\].
Now, we will do prime factorization of 120.
We can write 120 as a product of 2 numbers. Therefore,
\[12 = 4 \times 3\]
Breaking 4 into product of 2 numbers, we get
\[ \Rightarrow 12 = 2 \times 2 \times 3\]
So, we can write 12 as \[12 = {2^2} \times {3^1} \times {5^0}\]
Now, by using the factors, we will find the H.C.F of 120 and 12.
We know that the Highest Common Factors is the smallest power of the common factors.
The common factors of 120 and 12 are
\[120 = {2^3} \times {3^1}\]
\[12 = {2^2} \times {3^1}\]
Thus, HCF of 120 and 12 \[ = {2^2} \times {3^1}\]
Applying the exponent on terms, we get
\[ \Rightarrow \] HCF of (120, 12)\[ = 4 \times 3 = 12\]
\[ \Rightarrow \] HCF of \[\left( {\dfrac{{120}}{{100}},\dfrac{{12}}{{100}}} \right) = \dfrac{{12}}{{100}}\]
\[ \Rightarrow \] HCF of \[\left( {1.20,0.12} \right) = 0.12\]
Therefore, we can find the HCF of two given decimal numbers if there exists the equal number of decimal digits and the HCF of \[\left( {1.20,0.12} \right) = 0.12\]
Note:
We know that the Highest Common Factor is defined as the number which exactly divides all the numbers. Common Factor is the factor that is common to all the numbers whereas the prime factors are the factors, which is the product of the powers of the prime numbers. . Prime Factorization is a method of finding the factors of the given numbers. We should also make the decimal digits equal and then represent it in the form of fractions to find the Highest Common Factor.
Here, we will first convert the decimal numbers into fractions with the same denominator. Then we will find the factors of the numerators of both the fractions by using the Prime Factorization method. Then by using the common factors, we will find the HCF of both the decimal numbers.
Complete step by step solution:
We are given the numbers \[1.2\] and \[0.12\].
First, we will convert the numbers with equal numbers of decimal digits.
So, we get \[1.20\] and \[0.12\].
Now, we will convert both the decimal numbers in the form of fraction. So, we get
\[1.20 = \dfrac{{120}}{{100}}\]
\[0.12 = \dfrac{{12}}{{100}}\]
Since both the denominators are equal, we will find the HCF of the two numbers in the numerator 120 and 12.
We will find the factors of the numerators by using the prime factorization method.
First, we will do prime factorization of 120.
We can write 120 as a product of 2 numbers. Therefore,
\[120 = 12 \times 10\]
Breaking the numbers into product of 2 numbers, we get
\[\begin{array}{l} \Rightarrow 120 = 4 \times 3 \times 5 \times 2\\ \Rightarrow 120 = 2 \times 2 \times 2 \times 3 \times 5\end{array}\]
So, we can write 120 as \[120 = {2^3} \times {3^1} \times {5^1}\].
Now, we will do prime factorization of 120.
We can write 120 as a product of 2 numbers. Therefore,
\[12 = 4 \times 3\]
Breaking 4 into product of 2 numbers, we get
\[ \Rightarrow 12 = 2 \times 2 \times 3\]
So, we can write 12 as \[12 = {2^2} \times {3^1} \times {5^0}\]
Now, by using the factors, we will find the H.C.F of 120 and 12.
We know that the Highest Common Factors is the smallest power of the common factors.
The common factors of 120 and 12 are
\[120 = {2^3} \times {3^1}\]
\[12 = {2^2} \times {3^1}\]
Thus, HCF of 120 and 12 \[ = {2^2} \times {3^1}\]
Applying the exponent on terms, we get
\[ \Rightarrow \] HCF of (120, 12)\[ = 4 \times 3 = 12\]
\[ \Rightarrow \] HCF of \[\left( {\dfrac{{120}}{{100}},\dfrac{{12}}{{100}}} \right) = \dfrac{{12}}{{100}}\]
\[ \Rightarrow \] HCF of \[\left( {1.20,0.12} \right) = 0.12\]
Therefore, we can find the HCF of two given decimal numbers if there exists the equal number of decimal digits and the HCF of \[\left( {1.20,0.12} \right) = 0.12\]
Note:
We know that the Highest Common Factor is defined as the number which exactly divides all the numbers. Common Factor is the factor that is common to all the numbers whereas the prime factors are the factors, which is the product of the powers of the prime numbers. . Prime Factorization is a method of finding the factors of the given numbers. We should also make the decimal digits equal and then represent it in the form of fractions to find the Highest Common Factor.
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