Use extended Euclidean algorithm to find $j,k$ such that $52j+15k=3$ Is this questions suggesting that $\gcd(52,15)=3$ i.e. $52j+15k=3$?
if it is then why am I getting $1$ when I am computing the $\gcd$.
$$
\gcd(52,15)
= \gcd(15,7)
= \gcd(7,1)
= \gcd(1,0)
= 1
$$
Am I going in the right direction or not?
 A: First, we find $\gcd(\color{#c00}{52}, \color{#0a0}{15})$ using the Euclidean Algorithm.
$$
\begin{align*}
\color{#c00}{52}&=3 * \color{#0a0}{15} + \color{#00c}{7}\\
\color{#0a0}{15}&=2 * \color{#00c}{7} + 1\\
\color{#00c}{7}&=1*\color{#00c}{7}+0
\end{align*}
$$
So, $\gcd(\color{#c00}{52}, \color{#0a0}{15})=1$. You correctly calculated the $\gcd$ to be one. Now we rewrite the previous equations in terms of the remainders so we can substitute them back in:
$$
\begin{align*}
\color{#00c}{7}&=\color{#c00}{52}-3*\color{#0a0}{15}\\
1&=\color{#0a0}{15}-2*\color{#00c}{7}
\end{align*}
$$
which gives
$$
\begin{align*}
1&=\color{#0a0}{15}-2*(\color{#c00}{52}-3*\color{#0a0}{15})\\
1&=\color{#0a0}{15}-2*\color{#c00}{52}+6*\color{#0a0}{15}\\
1&=\color{#00c}{7}*\color{#0a0}{15}-2*\color{#c00}{52}.
\end{align*}
$$
To finish, we multiply the whole equation by $3$:
$$
\begin{align*}
3(1)&=3(\color{#00c}{7}*\color{#0a0}{15}-2*\color{#c00}{52})\\
3&=3*\color{#00c}{7}*\color{#0a0}{15}-3*2*\color{#c00}{52}\\
3&=21*\color{#0a0}{15}-6*\color{#c00}{52}.
\end{align*}
$$
Thus, we have that $j=-6$ and $k=21$.
