Minimize $152207x-81103y$ over the positive integers.

Since $\gcd (152207,81103)=1111$ it is the same as minimum of $$1111(137x-73y)$$

Since $137x-73y=1$ is solvable (say $x=8$ and $y=15$) the answer is $1111$.

Note $c=ax+by$ has integer solution if and only if $gcd(a,b)|c$ and if this exists then infinite no of integer solutions can be obtained from 1 solution by

$x=x_{0}+\frac{b}{d}k$ and $y=y_{0}-\frac{a}{d}k$

where $d$ is the gcd. And $x_{0},y_{0}$ are one solution which can be obtained from euclid's algorithm. (As will jagy has mentioned)

And also the defination of $gcd(a,b)$ is least positive value of the $ax+by=d$ where $x,y$ any integer.

$$ \gcd( 152207, 81103 ) = ??? $$

$$ \frac{ 152207 }{ 81103 } = 1 + \frac{ 71104 }{ 81103 } $$ $$ \frac{ 81103 }{ 71104 } = 1 + \frac{ 9999 }{ 71104 } $$ $$ \frac{ 71104 }{ 9999 } = 7 + \frac{ 1111 }{ 9999 } $$ $$ \frac{ 9999 }{ 1111 } = 9 + \frac{ 0 }{ 1111 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccccccc} & & 1 & & 1 & & 7 & & 9 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 2 }{ 1 } & & \frac{ 15 }{ 8 } & & \frac{ 137 }{ 73 } \end{array} $$ $$ $$ $$ 137 \cdot 8 - 73 \cdot 15 = 1 $$

$$ \gcd( 152207, 81103 ) = 1111 $$
$$ 152207 \cdot 8 - 81103 \cdot 15 = 1111 $$