Question

# Find the HCF by successive division method: 567, 621, 675.

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Hint: We will first understand the concept of successive division method which is given as if a quotient of the dividend is taken and this is used as the dividend in the next division, such division is called “successive division”. We will solve this in two phases i.e. first we will find HCF of 567, 621. On getting an answer, we will find HCF of that answer, 675. On getting an answer to this, we will find the HCF of 567, 621, 675.

Here, we will first understand the concept of successive division method.
If a quotient of the dividend is taken and this is used as the dividend in the next division, such division is called “successive division”. This process can continue up to any number of steps until the quotient in the first division is taken and dividend in the second division; the quotient in the second division is taken as the dividend in the third division; the quotient in the third division is taken as the dividend in the fourth division and so on.
Now, we solve this in two phases. First, we will take 567, 621.
So, now we will divide 621 by 567. We get as
567\overset{1}{\overline{\left){\begin{align} & 621 \\ & 567 \\ & \overline{\text{ }54} \\ \end{align}}\right.}}
Now, we will take divisor 567 as dividend and remainder 54 as divisor. On solving, we get as
54\overset{10}{\overline{\left){\begin{align} & 567 \\ & 540 \\ & \overline{\text{ 27}} \\ \end{align}}\right.}}
Again, we will take divisor 54 as dividend and remainder as divisor. We will get as
27\overset{2}{\overline{\left){\begin{align} & 54 \\ & 54 \\ & \overline{\text{ 0}} \\ \end{align}}\right.}}
Now, we got remainder 0 so, we will consider the divisor to be HCF of 567, 621 i.e. 27.
Now, we will find HCF of HCF $\left( 567,621 \right)$ , 675 i.e. 27, 675.
Using the same concept done above, we will solve.
27\overset{24}{\overline{\left){\begin{align} & 675 \\ & 54\downarrow \\ & \overline{\text{135}} \\ & 132 \\ & \overline{\text{ }3} \\ \end{align}}\right.}}
Now, we will take divisor 27 as dividend and remainder 3 to be divisor. We get as
3\overset{9}{\overline{\left){\begin{align} & 27 \\ & 27 \\ & \overline{\text{ }0} \\ \end{align}}\right.}}
Thus, we can now say that divisor 3 is HCF of 567, 621, 675.
Thus, $HCF\left( 567,621,675 \right)$ is 3.

Note: Do not assume a successive division method to be the same as normal division we do in prime factorization. Though the answer will be correct, the method of solving will be wrong. So, students should be clear with the methods of finding the highest common factor and then solve as per asked in question.