2013 AMC 10A Problem 13

1. What the numbers look like. The first and last digits are the same, so the number is $a\,b\,a$, like $373$ or $808$. Here $a$ is $1$ to $9$ and $b$ is $0$ to $9$.
2. Not divisible by 5. A number is divisible by $5$ when it ends in $0$ or $5$. It ends in $a$, and $a$ is never $0$, so we just need $a \ne 5$.
3. Digit sum less than 20. The digits add to $a + b + a = 2a + b$, and we need $2a + b < 20$, so $b < 20 - 2a$.
4. Go through the values of $a$.

For small $a$ the biggest possible sum is $2 \times 4 + 9 = 17$, which is already under $20$, so every $b$ works.
5. Add. $40 + 8 + 6 + 4 + 2 = 60$.

The answer is $\boxed{\textbf{(B)}\ 60}$.

The "first digit equals last digit" rule leaves only two digits to choose. Each other rule then limits one of them. Going through $a$ one value at a time makes the digit-sum rule easy: it never matters for small $a$, and for large $a$ the counts drop by $2$ each time.

Count the bad ones and subtract. There are $9 \times 10 = 90$ numbers of the form $a\,b\,a$. Take away the $10$ with $a = 5$, leaving $80$. The ones with digit sum $20$ or more need $a \ge 6$: that is $2, 4, 6, 8$ numbers for $a = 6, 7, 8, 9$, or $20$ in all. So $80 - 20 = 60$.

Write the number as $\overline{aba}$ with $1 \le a \le 9$ and $0 \le b \le 9$. The last digit is $a$, so "not divisible by $5$" means $a \ne 5$ (and $a \ne 0$, already excluded). The digit sum condition is $2a + b < 20$.

Case $a \in \{1, 2, 3, 4\}$: $2a + b \le 8 + 9 = 17 < 20$, so all $10$ values of $b$ work: $4 \cdot 10 = 40$ numbers.

Case $a = 6$: need $b < 8$, so $b = 0, \dots, 7$: $8$ numbers.

Case $a = 7$: need $b < 6$: $6$ numbers.

Case $a = 8$: need $b < 4$: $4$ numbers.

Case $a = 9$: need $b < 2$: $2$ numbers.

Total: $40 + 8 + 6 + 4 + 2 = 60$.

The answer is $\boxed{\textbf{(B)}\ 60}$.