Question

Find the HCF by continued division method.
\[\begin{align}
  & A.60and95 \\
 & B.96and150 \\
 & C.575,75and290 \\
 & D.90,120and980 \\
\end{align}\]

Answer 1

Hint: In order to find the HCF by continued division method, we must consider the numbers at first. Then consider the smaller number among them and divide it with the other. Then we obtain a remainder. Now this remainder should be divided into the previously considered divisor and this process should continue until the obtained remainder cannot be divided further. The last divisor that divides will be the HCF of the numbers.

Complete step by step answer:
Now let us have a brief HCF of numbers. HCF is called the Highest Common Factor which means that a number which is common and highest divides the numbers. HCF is also known as the Greatest Common Divisor. The HCF of co-primes is always one. The HCF of two fractions is nothing but the HCF of numerators divided by LCM of denominators. We can find HCF of numbers in three ways. They are : listing out the common factors, the prime factorization method and the division method.
Now let us find the HCF of given numbers.
Let us consider \[A.60and95\]
\[\begin{align}
  & \text{60}\overline{\left){95}\right.}\left( 1 \right. \\
 & \,\,\,\,-60 \\
 & \,\,\,\,\,\,\,\,\,35\overline{\left){\text{60}}\right.}\left( \text{1} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,-35 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,25\overline{\left){\text{35}}\right.}\left( \text{1} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-25 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10\overline{\left){\text{25}}\right.}\left( \text{2} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-20 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\overline{\left){\text{10}}\right.}\left( \text{2} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-10} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
\end{align}\]
We can see that \[5\] is the last divisor in the above division process.
\[\therefore \] \[5\] is the HCF of \[60and95\].
Now, consider \[96and150\].

\[\begin{align}
  & \text{96}\overline{\left){150}\right.}\left( 1 \right. \\
 & \,\,\,\,\,\,\,-96 \\
 & \,\,\,\,\,\,\,\,\,\,\text{ 54}\overline{\left){\text{96}}\right.}\left( \text{1} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-54 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 42}\overline{\left){\text{54}}\right.}\left( \text{1} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-42 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 12}\overline{\left){\text{42}}\right.}\left( \text{3} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-36 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 6}\overline{\left){\text{12}}\right.}\left( \text{2} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-10} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}\]
We can see that \[\text{6}\] is the last divisor in the above division process.
\[\therefore \] \[\text{6}\] is the HCF of \[96and150\]
Now let us consider \[575,75and290\]
Since we have three numbers, we will be finding HCF of any two numbers first.
We get,
\[\begin{align}
  & 75\overline{\left){575}\right.}\left( 1 \right. \\
 & \,\,\,\,-525 \\
 & \,\,\,\,\,\,\,\,\,\,\text{ 50}\overline{\left){75}\right.}\left( \text{1} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-50 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 25}\overline{\left){\text{50}}\right.}\left( 2 \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-50} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}\]
The HCF of \[575,75\]is \[\text{25}\].
Now, with this HCF, we will be dividing the remaining number. We get,
\[\begin{align}
  & 25\overline{\left){290}\right.}\left( 11 \right. \\
 & \,\,\,\,-275 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,15\overline{\left){25}\right.}\left( \text{1} \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 10\overline{\left){\text{15}}\right.}\left( 1 \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-10 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\overline{\left){10}\right.}\left( 2 \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-10} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}\]
\[\therefore \] \[5\] is the HCF.
Now let us consider \[90,120and980\].
\[\begin{align}
  & 120\overline{\left){980}\right.}\left( 8 \right. \\
 & \,\,\,\,\,\,\,-960 \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\text{ 20}\overline{\left){120}\right.}\left( 6 \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-120} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}\]
Now with\[\text{20}\] as a divisor, we will be dividing the third number.
\[\begin{align}
  & 20\overline{\left){90}\right.}\left( 4 \right. \\
 & \,\,\,\,\,\text{ -80} \\
 & \,\,\,\,\,\,\,\text{ 10}\overline{\left){20}\right.}\left( 2 \right. \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\underline{-20} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
\end{align}\]
\[\therefore \] \[\text{10}\] is the HCF.

Note: We must always have a note while following the division method, only two numbers are to be considered at first and then the HCF of the third number should only be considered as the divisor for the third number. The final divisor would be the HCF.