[Number Theory]
Problem: Let $N$ be the least integer greater than 20 that is a palindrome in both base 20 and base 14 . For example, the three-digit base-14 numeral (13)5(13) ${ }_{14}$ (representing $13 \cdot 14^{2}+5 \cdot 14^{1}+13 \cdot 14^{0}$ ) is a palindrome in base 14 , but not in base 20 , and the three-digit base-14 numeral (13)31 14 is not a palindrome in base 14 . Compute the base-10 representation of $N$.

Solution: Because $N$ is greater than 20, the base-20 and base-14 representations of $N$ must be at least two digits long. The smallest possible case is that $N$ is a two-digit palindrome in both bases. Then $N=20 a+a=21 a$, where $1 \leq a \leq 19$. Similarly, in order to be a two-digit palindrome in base $14, N=14 b+b=15 b$, with $1 \leq b \leq 13$. So $N$ would have to be a multiple of both 21 and 15 . The least common multiple of 21 and 15 is 105 , which has the base 20 representation of $105=55_{20}$ and the base-14 representation of $105=77_{14}$, both of which are palindromes. Thus the answer is 105.

Final answer: 105