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If $a,b$ are nonzero integers, $d=\gcd (a,b)$, and $(x_0,y_0)$ is one solution of the linear Diophantine equation $ax+by=c$, then all solutions areSubtract two solutions and use the fact that $a/d$ and $b/d$ are coprime integers. The linear Diophantine equation solvability criterion says that a particular solution exists exactly when $d$ divides $c$; the extended Euclidean algorithm constructs one. This parametrization separates existence from the complete family of answers.

Use the Euclidean algorithm: $23=18+5$, $18=3\cdot 5+3$, $5=3+2$, and $3=2+1$. Back-substitution gives the Bezout identity $1=9\cdot 18-7\cdot 23$. Multiplying by $101$ gives one solution of the linear Diophantine equation, and shifting the coefficients by multiples of $(23,-18)$ reduces it toIndeed, $18\cdot 12+23(-5)=216-115=101$. More generally all solutions are $x=12+23k$, $y=-5-18k$, $k\in \mathbb{Z}$: subtracting this particular solution gives $18(x-12)=-23(y+5)$, and the coprime integers $18,23$ force $23\mid (x-12)$.

The Euclidean algorithm givesThus the greatest common divisor is $d=7$. The extended Euclidean algorithm reverses these divisions:This supplies the Bezout identity with $x=8$, $y=-11$.

To obtain all solutions of a linear Diophantine equation, subtract this particular solution from any other and divide by $7$. The result is $29(x-8)+21(y+11)=0$. Since $29$ and $21$ are coprime integers, $21$ divides $x-8$, givingEvery displayed pair satisfies the original linear Diophantine equation, so the parametrization is exhaustive.

For the modular arithmetic calculation, $21\cdot 18=378\equiv 1(\mathrm{mod}29)$, so $18$ is the modular inverse of $21$. Multiplication gives $n\equiv 25\cdot 18\equiv 15(\mathrm{mod}29)$. The admissible integers are $15+29t$ for $0\le t\le 68$, ending at $1987$; the next is $2016$. There are $\boxed{69}$ such integers.