Find the HCF using division method \[391,425,527\]
Answer
581.7k+ views
Hint:To find the HCF using the division method, we divide the first two terms of the three terms and the first term of the three term acts as a divisor for the second term and the second term acts as a divisor for the third term and even after that if the remainder is zero then we get the final answer on the right side as the HCF of these three numbers.
Complete step by step solution:
To find the HCF of the three numbers as given in the above question we use the long division formula, first for the first two numbers i.e. \[391,425\].
Now let us make the value \[391\] as divisor and \[425\] as dividend, for which we get the value as:
\[\begin{align}
& \begin{matrix}
\left. 391 \right) & 425 \\
{} & \underline{391} \\
\end{matrix} \\
& \text{ }\begin{matrix}
\left. 34 \right) & \left. 391 \right)1 \\
{} & \underline{34}\text{ } \\
\end{matrix} \\
& \text{ 51} \\
& \text{ }\underline{\text{34}} \\
& \text{ }\left. \text{17} \right)\left. 34 \right)2 \\
& \text{ }\underline{\text{34}} \\
& \text{ 0} \\
\end{align}\]
We can see that the numbers \[391,425\] cleanly divides both the number forming the HCF of the two numbers as \[17\] and now we follow the same procedure for the number \[425,527\].
Now let us make the value \[425\] as divisor and \[527\] as dividend, for which we get the value as:
\[\begin{align}
& \begin{matrix}
\left. 425 \right) & 527 \\
{} & \underline{425} \\
\end{matrix} \\
& \text{ }\begin{matrix}
\left. 17 \right) & \left. 17 \right)1 \\
{} & \text{ }17\text{ } \\
\end{matrix} \\
& \text{ 0} \\
& \text{ } \\
\end{align}\]
Now for the division of \[425\] and \[527\], we get the HCF as \[17\].
Therefore, the two HCF are \[17\] making the highest common factor as \[17\].
Note: The HCF can also be found by the factorization method where we use the numbers as:
\[\begin{align}
& 17\left| 391,425,527 \right. \\
& \text{ }\left| 023,025,031 \right. \\
\end{align}\]
Hence, the HCF of the three numbers \[391,425,527\] is \[17\].
Complete step by step solution:
To find the HCF of the three numbers as given in the above question we use the long division formula, first for the first two numbers i.e. \[391,425\].
Now let us make the value \[391\] as divisor and \[425\] as dividend, for which we get the value as:
\[\begin{align}
& \begin{matrix}
\left. 391 \right) & 425 \\
{} & \underline{391} \\
\end{matrix} \\
& \text{ }\begin{matrix}
\left. 34 \right) & \left. 391 \right)1 \\
{} & \underline{34}\text{ } \\
\end{matrix} \\
& \text{ 51} \\
& \text{ }\underline{\text{34}} \\
& \text{ }\left. \text{17} \right)\left. 34 \right)2 \\
& \text{ }\underline{\text{34}} \\
& \text{ 0} \\
\end{align}\]
We can see that the numbers \[391,425\] cleanly divides both the number forming the HCF of the two numbers as \[17\] and now we follow the same procedure for the number \[425,527\].
Now let us make the value \[425\] as divisor and \[527\] as dividend, for which we get the value as:
\[\begin{align}
& \begin{matrix}
\left. 425 \right) & 527 \\
{} & \underline{425} \\
\end{matrix} \\
& \text{ }\begin{matrix}
\left. 17 \right) & \left. 17 \right)1 \\
{} & \text{ }17\text{ } \\
\end{matrix} \\
& \text{ 0} \\
& \text{ } \\
\end{align}\]
Now for the division of \[425\] and \[527\], we get the HCF as \[17\].
Therefore, the two HCF are \[17\] making the highest common factor as \[17\].
Note: The HCF can also be found by the factorization method where we use the numbers as:
\[\begin{align}
& 17\left| 391,425,527 \right. \\
& \text{ }\left| 023,025,031 \right. \\
\end{align}\]
Hence, the HCF of the three numbers \[391,425,527\] is \[17\].
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
How many of the following diseases can be studied with class 11 biology CBSE

Which of the following enzymes is used for carboxylation class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Receptor sites for neurotransmitters are present on class 11 biology CBSE

JGA JuxtaGlomerular Apparatus a sensitive region which class 11 biology CBSE

Root hairs develop from the region of A Meristematic class 11 biology CBSE

