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  • Find the HCF of 65 and 117 and express it in the form 65m+117n.

Find the HCF of 65 and 117 and express it in the form 65m+117n.

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Given integers are 65 and 117 such that 117>65
 

Applying division lemma to 65 and 117, we get
117=65×1+52
Since the remainder 52\neq0. So, 
65=52×1+13  

We consider the new divisor 52 and the new remainder 13 , then we get - 
52=13×4+0
 

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

\\ From\ $(i i),$ \ we \ have \ $13=65-52 \times 1$ \\ $\Rightarrow \quad 13=65-(117-65 \times 1)$ \\ $\Rightarrow \quad 13=65-117+65 \times 1$ \\ $\Rightarrow \quad 13=65 \times 2+117 \times(-1)$ \\ $\Rightarrow \quad 13=65 m+117 n,$ \ where \ $m=2$ \ and \ $n=-1$

Posted by

Deependra Verma

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