If the HCF of 65 and 117 is expressed in the form of $65n-117$. Find the value of $n$.
Answer
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Hint: We need to find the HCF of 65 and 117. First, we need to find the common factors of 117 and 65 from their factors’ list. Then we find the greatest common factor of 117 and 65. We can also take the simultaneous factorisation of those two numbers to find the HCF.
Complete step by step solution:
We need to find the HCF of 65 and 117. HCF stands for the highest common factor.
We use the simultaneous factorisation to find the highest common factor of 117 and 65.
We have to divide both of them with possible primes which can divide both of them.
\[\begin{align}
& 13\left| \!{\underline {\,
65,117 \,}} \right. \\
& 1\left| \!{\underline {\,
5,9 \,}} \right. \\
\end{align}\]
The only possible prime being 13. Therefore, the highest common factor of 117 and 65 is 13.
It is given that the HCF of 65 and 117 is expressed in the form of $65n-117$.
\[\begin{align}
& 65n-117=13 \\
& \Rightarrow 65n=117+13=130 \\
& \Rightarrow n=\dfrac{130}{65}=2 \\
\end{align}\]
Therefore, the value of $n$ is 2.
Note:
We need to remember that the HCF has to be only one number. It is the highest possible divisor of all the given numbers. If the given numbers are prime numbers, then the HCF of those numbers will always be 1.
Therefore, if for numbers $x$ and $y$, the HCF is $a$ then the HCF of the numbers $\dfrac{x}{a}$ and $\dfrac{y}{a}$ will be 1.
Complete step by step solution:
We need to find the HCF of 65 and 117. HCF stands for the highest common factor.
We use the simultaneous factorisation to find the highest common factor of 117 and 65.
We have to divide both of them with possible primes which can divide both of them.
\[\begin{align}
& 13\left| \!{\underline {\,
65,117 \,}} \right. \\
& 1\left| \!{\underline {\,
5,9 \,}} \right. \\
\end{align}\]
The only possible prime being 13. Therefore, the highest common factor of 117 and 65 is 13.
It is given that the HCF of 65 and 117 is expressed in the form of $65n-117$.
\[\begin{align}
& 65n-117=13 \\
& \Rightarrow 65n=117+13=130 \\
& \Rightarrow n=\dfrac{130}{65}=2 \\
\end{align}\]
Therefore, the value of $n$ is 2.
Note:
We need to remember that the HCF has to be only one number. It is the highest possible divisor of all the given numbers. If the given numbers are prime numbers, then the HCF of those numbers will always be 1.
Therefore, if for numbers $x$ and $y$, the HCF is $a$ then the HCF of the numbers $\dfrac{x}{a}$ and $\dfrac{y}{a}$ will be 1.
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