Two sides of an integer sided triangle have lengths $$18$$ and $$x$$ where $$x<100$$. If there are exactly $$35$$ possible integer values $$y$$ such that $$18, x, y$$ are the sides of a non-degenerate triangle, find the number of possible integer values $$x$$ can have.

The three sides must satisfy the strict triangle inequalities (non-degenerate triangle):

$$18 + x \gt y,\qquad 18 + y \gt x,\qquad x + y \gt 18$$

Combining the first and last of these gives one pair of two-sided bounds for $$y$$:

$$|18 - x| \lt y \lt 18 + x$$

Because $$y$$ is an integer, the smallest admissible integer is $$|18-x|+1$$ and the largest is $$18+x-1$$.
Hence the total number of integer values of $$y$$ possible for a given $$x$$ is

$$N(x)=\bigl(18+x-1\bigr)-\bigl(|18-x|+1\bigr)+1 =17+x-|18-x|$$

The problem states that exactly $$35$$ such integers $$y$$ exist, so

$$17+x-|18-x|=35 \;\;\Longrightarrow\;\; x-|18-x|=18$$

Case 1: $$x\ge 18$$
Then $$|18-x|=x-18$$, giving $$x-(x-18)=18\;\Longrightarrow\;18=18$$, which is always true. Thus every integer $$x\ge 18$$ yields $$N(x)=35$$.

Case 2: $$x\le 17$$
Here $$|18-x|=18-x$$, so $$x-(18-x)=18\;\Longrightarrow\;2x-18=18\;\Longrightarrow\;x=18,$$ which contradicts $$x\le17$$. Therefore no $$x\le17$$ works.

Because the question restricts to $$x\lt 100$$, the valid integers are $$x=18,19,\dots ,99$$.

The count of these integers is $$99-18+1=82$$.

Hence, the number of possible integer values of $$x$$ is 82.