Problem (Number Theory):
A palindrome is a positive integer, not ending in 0 , that reads the same forwards and backwards. For example, 35253,171,44, and 2 are all palindromes, but 17 and 1210 are not. Compute the least positive integer greater than 2013 that cannot be written as the sum of two palindromes.

Solution:
If $a+b \geq 2014$, then at least one of $a, b$ must be greater than 1006 . The palindromes greater than 1006 but less than 2014 are, in descending order, 2002, 1991, 1881, ..., 1111. Let a



represent the larger of the two palindromes. Then for $n=2014, a=2002$ is impossible, because $2014-2002=12$. Any value of $a$ between 1111 and 2000 ends in 1 , so if $a+b=2014$, $b$ ends in 3 , and because $b<1000$, it follows that $303 \leq b \leq 393$. Subtracting 303 from 2014 yields 1711, and so $a \leq 1711$. Thus $a=1661$ and $b=353$. A similar analysis shows the following results:

$$
\begin{aligned}
& 2015=1551+464 ; \\
& 2016=1441+575 ; \\
& 2017=1331+686 ; \text { and } \\
& 2018=1221+797
\end{aligned}
$$

But 2019 cannot be expressed as the sum of two palindromes: $b$ would have to end in 8 , so $b=808+10 d$ for some digit $d$. Then $2019-898 \leq a \leq 2019-808$, hence $1121 \leq a \leq 1211$, and there is no palindrome in that interval.

Final answer: 2019