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Find the HCF using division method \[391,425,527\]

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Last updated date: 27th Feb 2024
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IVSAT 2024
Answer
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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 \[17\].