Other people have have answered. But the thing to take away from this is the idea of Bezout's Lemma (sometimes known as Bezout's Identity).
If $M,N$ are integers with greatest common divisor $\gcd(M,N)$ then there will exist integers $a,b$ so that $Ma + Nb = \gcd(M,N)$.
Another way of putting this is
If $j,k$ are relatively prime integers, then there will exist integers $a,b$ so that $ja + kb = 1$.
If we note that $\gcd(M,N)|M$ and $\gcd(M,N)$ then $\gcd(M,N)|Ma + Nb$ for any integers $a,b$ which leads to a third way of putting this
(Version 3) If $M,N$ are integers then:
For any integers $a,b$ the $Ma + Nb$ will be a multiple of $\gcd(M,N)$.
Integers $c,d$ exist so that $Mc + Nd = \gcd(M,N)$.
and therefore
- For any multiple of $\gcd(M,N)$, say $k\gcd(M,N)$ for some integer $k$, then integers $a,b$ exists so that $Ma + Nb = k\gcd(M,N)$. (Just let $a = kc; b=kd$ where $c,d$ are as in 2. above.)
And this answers your question.
$152207x−81103y$ will always be a multiple of $\gcd(152207, 81103) = 1111$
$152207x - 81103y = 1111$ will be possible.
And as the smallest positive integer that is a multiple of $1111$ is $1111$, the smallest positive value of $152207x -88103y$ is $1111$.
Notice, we don't have to actually find the values that make this true. It's enough to know it can be done!
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Addendum:
1) Note: I never said and never implied that any of those integer pairs were unique.
$Ma + Nb = k\gcd(M,N)$ will actually have an infinite number of solutions.
Notice $M(a \pm w\frac N{\gcd(M,N}) + N(b \mp w\frac M{\gcd(M,N)}) = Ma + Nb = k\gcd(M,N)$ will always be a solution. But all solutions will be is such a form.
2) To actually find a solution
$152207x -88103y =1111$ we can use Euclid Algorithm
$152207 = 81103 + 71104; 71104 = 152207 - 81103$
$81103 = 71104 + 9999; 9999 = 81103 - 71104 = 81103 -(152207-81103) = 2*81103-152207$
$71104 = 7*9999 + 1111; 1111 = 71104 - 7*9999=(152207 - 81103)-7(2*81103-152207)=8*152207- 15*81103$
$9999 = 9*1111 + 0$ thats as far as we can go.
So for $x = 8; y = -15$ we get $152207x+81103y = \gcd(152207, 81103)$.