Question

if the hcf of 657 and 963 is expressible in form of 657x + 963 x (-15) find x

Answer 1

Solution:

Frist we need to find the H.C.F of 657 and 963

It would be multiple by one more that is 9 .

As ,

657 = 9×63

         963 = 9×107

As we know , the H.C.F should be Express in the form of 657x + 963 x (-15).

For Finding Vlaue :

657x +  963×(-15) = 9

=> 657x - 963 ×15 = 9

=> 657x - 14445 = 9

=> 657x = 14445 + 9

=> 657x = 14454

$x = \frac{14454}{657}$

=> x = 22

So, the value of x is 22

Answer 2

S O L U T I O N :

$\bigstar$ Firstly, we find H.C.F. (Highest Common Factor):

657 & 963

$\begin{array}{r|l} 3 & 657 \\ \cline{2-2} 3 & 219 \\ \cline{2-2} 73 & 73 \\ \cline{2-2} & 1\end{array}$

&

$\begin{array}{r|l} 3 & 963 \\ \cline{2-2} 3 & 321 \\ \cline{2-2} 107 & 107\\ \cline{2-2} & 1\end{array}$

The prime factorization of 657 = $\sf{\boxed{3}\times \boxed{3} \times 73}}$

The prime factorization of 963 = $\sf{\boxed{3}\times \boxed{3} \times 107}}$

∴ H.C.F . we get = 3 × 3 = 9

A/q

$\longrightarrow\tt{657x+963\times (-15)=9}\\\\\longrightarrow\tt{657x+(-14445)=9}\\\\\longrightarrow\tt{657x-14445=9}\\\\\longrightarrow\tt{657x=9+14445}\\\\\longrightarrow\tt{657x=14454}\\\\\longrightarrow\tt{x=\cancel{14454/657}}\\\\\longrightarrow\bf{x=22}$

Thus;

$\underbrace{\sf{The\:value \:of\:x\:=\:22}}}$