Question

Find the HCF of 65 and 117 and express it in the form 65 m + 117n. [4 MARKS]

Answer 1

Concept : 1 Mark
Application: 2 Marks
Representation : 1 Mark

Given integers are 65 and 117 and 117 > 65

Applying division lemma to 65 and 117, we get

$\begin{array}{ccc}65& 117& 1\\ & 65& \\ & 52& \end{array}$

$117=65\times 1+52$ ......... (i)

Since the remainder $52\ne 0$. So, we apply the division lemma to the divisor 65 and the remainder 52 to get

$\begin{array}{ccc}52& 65& 1\\ & 52& \\ & 13& \end{array}$

$65=52\times 1+13$ .......... (ii)

We consider the new divisor 52 and the new remainder 13 and apply division lemma, to get

$52=13\times 4+0$ .......... (iii)

At this stage the remainder is zero. So, the last divisor or the non-zero remainder at the earlier stage i.e. 13 is the HCF of 65 and 117.

From (ii), we have

$13=65-52\times 1$

$\Rightarrow 13=65-(117-65\times 1)$
[Substituting $52=117-65\times 1$ obtain from (i)]

$\Rightarrow 13=65-117+65\times 1$

$\Rightarrow 13=65\times 2+117\times (-1)$

$\Rightarrow 13=65m+117n$, where m = 2 and n = -1