Answer
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Hint:Here we will find the value of H.C.F by using the prime factorization method. In the prime factorization method, we find the prime factor of all the numbers and then multiply all the common factors in them to get our H.C.F.
Complete step-by-step answer:
We will use the prime factorization method to find the H.C.F of 150, 140, and 210.
Firstly, we will write all the numbers as a product of their prime factors:
Starting with 150 we will divide 150 by each prime number starting with 2 and keep on dividing it till we get 1 as shown below:
$\begin{array}{*{20}{l}}
2\| \underline {150} \\
\hline
3\| \underline {75} \\
\hline
5\| \underline {25} \\
\hline
5\| \underline 5 \\
\hline
{}\| \underline 1
\end{array}$
Therefore, the prime factors of 150 are:
\[150 = 2 \times 3 \times 5 \times 5\]……\[\left( 1 \right)\]
Next, we will find the prime factor of 140 as shown below:
$\begin{array}{*{20}{l}}
2\| \underline {140} \\
\hline
2\| \underline {70} \\
\hline
5\| \underline {35} \\
\hline
7\| \underline 7 \\
\hline
{}\| \underline 1
\end{array}$
Therefore, the prime factors of 140 are:
\[140 = 2 \times 2 \times 5 \times 7\]…….\[\left( 2 \right)\]
Next, we will find prime factor of 210 as shown below:
$\begin{array}{*{20}{l}}
2\| \underline {210} \\
\hline
3\| \underline {105} \\
\hline
5\| \underline {35} \\
\hline
7\| \underline 7 \\
\hline
{}\| \underline 1
\end{array}$
Therefore, the prime factors of 210 are:
\[210 = 2 \times 3 \times 5 \times 7\]…….\[\left( 3 \right)\]
From the equation \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] we get the common factors as \[2 \times 5\]. Therefore,
\[H.C.F\left( {150,140,210} \right) = 2 \times 5\]
Multiplying the terms, we get
\[ \Rightarrow H.C.F\left( {150,140,210} \right) = 10\]
Note: H.C.F (Highest Common Factor) is the greatest number that divides each of the given integers completely. H.C.F contains all the common factors of two numbers and thus we can say H.C.F of a co-prime number is always equal to 1 as they won’t have any common divisor. The product of two numbers is equal to the product of their H.C.F and L.C.M.
Complete step-by-step answer:
We will use the prime factorization method to find the H.C.F of 150, 140, and 210.
Firstly, we will write all the numbers as a product of their prime factors:
Starting with 150 we will divide 150 by each prime number starting with 2 and keep on dividing it till we get 1 as shown below:
$\begin{array}{*{20}{l}}
2\| \underline {150} \\
\hline
3\| \underline {75} \\
\hline
5\| \underline {25} \\
\hline
5\| \underline 5 \\
\hline
{}\| \underline 1
\end{array}$
Therefore, the prime factors of 150 are:
\[150 = 2 \times 3 \times 5 \times 5\]……\[\left( 1 \right)\]
Next, we will find the prime factor of 140 as shown below:
$\begin{array}{*{20}{l}}
2\| \underline {140} \\
\hline
2\| \underline {70} \\
\hline
5\| \underline {35} \\
\hline
7\| \underline 7 \\
\hline
{}\| \underline 1
\end{array}$
Therefore, the prime factors of 140 are:
\[140 = 2 \times 2 \times 5 \times 7\]…….\[\left( 2 \right)\]
Next, we will find prime factor of 210 as shown below:
$\begin{array}{*{20}{l}}
2\| \underline {210} \\
\hline
3\| \underline {105} \\
\hline
5\| \underline {35} \\
\hline
7\| \underline 7 \\
\hline
{}\| \underline 1
\end{array}$
Therefore, the prime factors of 210 are:
\[210 = 2 \times 3 \times 5 \times 7\]…….\[\left( 3 \right)\]
From the equation \[\left( 1 \right)\],\[\left( 2 \right)\] and \[\left( 3 \right)\] we get the common factors as \[2 \times 5\]. Therefore,
\[H.C.F\left( {150,140,210} \right) = 2 \times 5\]
Multiplying the terms, we get
\[ \Rightarrow H.C.F\left( {150,140,210} \right) = 10\]
Note: H.C.F (Highest Common Factor) is the greatest number that divides each of the given integers completely. H.C.F contains all the common factors of two numbers and thus we can say H.C.F of a co-prime number is always equal to 1 as they won’t have any common divisor. The product of two numbers is equal to the product of their H.C.F and L.C.M.
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