2013 AMC 10A Problem 13

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 $\boxed{\textbf{(B)}\ 60}$.

The palindrome condition reduces the problem to two free digits, and each remaining condition constrains one of them. Handling the outer digit $a$ case by case keeps the digit-sum inequality simple: it is automatic for small $a$ and a short arithmetic progression $8, 6, 4, 2$ for large $a$.

Count the complement: there are $9 \cdot 10 = 90$ palindromes $\overline{aba}$. Remove the $10$ with $a = 5$, leaving $80$. Among those, digit sum $\ge 20$ requires $a \ge 6$ with $b \ge 20 - 2a$: that is $2 + 4 + 6 + 8 = 20$ numbers. $80 - 20 = 60$.

• Forgetting that 'not divisible by 5' rules out a = 5 as the last digit, which adds 10 extra numbers and gives 70.
• Using "sum less than 20" as "at most 20," which admits $b = 20 - 2a$ and adds $4$ numbers.