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I am not able to establish that the GCD will not depend upon the value of $n$. Although by taking some initial values of $n$, the GCD is always 1, how do we prove that it is always 1 whatever be the value of $n$.

Kings
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  • Let $a=22n+1$ and $b=33n+2$. What is $3a-2b$? – lhf Nov 22 '22 at 10:16
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1 Answers1

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Let us denote the numbers by $A$ and $B$, where

$A=22n+1$

$B=33n+2$

Our first task is to establish a relation between $A$ and $B$ such that $n$ is eliminated.

$3A=66n+3$ and $2B=66n+4$

$\Rightarrow 2B-3A=1$

Now suppose GCD of $A,\,\,B$ is $k$, then $2B-3A$should be a multiple of $k$, but we have just proved that $2B-3A=1$, therefore $k=1$ only.

Anil Sharma
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