Question

Today's interesting question is:-
If HCF of 65 and 117 is expressible in form of 65n-117,then find the value of n
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Answer 1

Step-by-step explanation:

The last non-zero remainder obtained was 13 which is the HCF of 65 and 117. On comparing it with 65n - 117 we get the value of n as 2.

Answer 2

❍Given:-

$65n - 117$

❍To Assume:-

Solve the equation

$65n - 117$

❍Solution:-

$117 = 1 \times 65 + 52 \\ 65 = 1 \times 52 + 13 \\ 52 = 4 \times 13 + 0 \\ hcf \: of \: 65 \: and \: 117 \: is \: 13 \\ \\ ∴65n - 117 = 13 \\ ∴65n = 117 + 13 = 130 \\ ∴n \frac{130}{65} \\ ∴n = 2✓$

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