jmc

Our goal is to count the two-digit integers in the form $8n+2$ for integer values of $n$. We examine the inequality, $$10\le 8n+2<100.$$Subtracting 2 from all parts simplifies things: $$8\le 8n<98.$$Dividing everything by 8 isolates the possible values of $n$: $$1\le n<12\frac{1}{4}.$$Since $n$ can be any integer from 1 to 12, there are $\boxed{12}$ two-digit integers in the form $8n+2$ (that leave a remainder of 2 when divided by 8).