Find the HCF of the following pair of integers and express it as a linear combination of them:
506 and 1155

Given: 506 and 1155

To do: Here we have to find the HCF of the given pair of integers and express it as a linear combination.


Solution:

Using Euclid's division algorithm to find HCF:

Using Euclid’s lemma to get: 
  • $1155\ =\ 506\ \times\ 2\ +\ 143$   ...(i)

Now, consider the divisor 506 and the remainder 143, and apply the division lemma to get:
  • $506\ =\ 143\ \times\ 3\ +\ 77$   ...(ii)

Now, consider the divisor 143 and the remainder 77, and apply the division lemma to get:
  • $143\ =\ 77\ \times\ 1\ +\ 66$   ...(iii)

Now, consider the divisor 77 and the remainder 66, and apply the division lemma to get:
  • $77\ =\ 66\ \times\ 1\ +\ 11$   ...(iv)

Now, consider the divisor 66 and the remainder 11, and apply the division lemma to get:
  • $66\ =\ 11\ \times\ 6\ +\ 0$   ...(v)

The remainder has become zero, and we cannot proceed any further. 

Therefore the HCF of 506 and 1155 is the divisor at this stage, i.e., 11.


Expressing the HCF as a linear combination of 506 and 1155:

$11\ =\ 77\ –\ 66\ \times\ 1$   {from equation (iv)}

$11\ =\ 77\ –\ [143\ –\ 77\ \times\ 1]\ \times\ 1$   {from equation (iii)}

$11\ =\ 77\ –\ 143\ +\ 77\ \times\ 1$

$11\ =\ 77\ \times\ 2\ –\ 143$

$11\ =\ [506\ –\ 143\ \times\ 3]\ \times\ 2\ –\ 143$   {from equation (ii)}

$11\ =\ 506\ \times\ 2\ –\ 143\ \times\ 6\ –\ 143$

$11\ =\ 506\ \times\ 2\ –\ 143\ \times\ 7$

$11\ =\ 506\ \times\ 2\ –\ [1155\ –\ 506\ \times\ 2]\ \times\ 7$   {from equation (i)}

$11\ =\ 506\ \times\ 2\ –\ 1155\ \times\ 7\ +\ 506\ \times\ 14$

$\mathbf{11\ =\ 506\ \times\ 16\ –\ 1155\ \times\ 7}$


So, HCF of 506 and 1155 is 11 and it can be expressed as $11\ =\ 506\ \times\ 16\ –\ 1155\ \times\ 7$.

Updated on: 10-Oct-2022

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