imc

We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$.($\overline{abc}$ denotes the number $100a+10b+c$). We see that $x\ne 0$ and $x\ne 5$, as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5. The second condition is that the sum $2x+y<20$. When $x$ is $1$, $2$, $3$, or $4$, $y$ can be any digit from $0$ to $9$, as $2x<10$. This yields $10(4)=40$ numbers. When $x=6$, we see that $12+y<20$ so $y<8$. This yields $8$ more numbers. When $x=7$, $14+y<20$ so $y<6$. This yields $6$ more numbers. When $x=8$, $16+y<20$ so $y<4$. This yields $4$ more numbers. When $x=9$, $18+y<20$ so $y<2$. This yields $2$ more numbers. Summing, we get $40+8+6+4+2=\boxed{60}$