Find the smaller palindrome that is divisible by $3,5,11$

Question

It seems to me that there are only two ways to solve the problem:
First-Use divisibility rules to select last digits, and work my way from there
Second-Write out all multiples of 165 since 3, 5, and 11 are relatively prime
Both ways seem pretty slow, what is the fast way to solve it?

Answer 1

How about $5115$. Here's how I got it:

• Divisibility by $5$, the only units digits divisible by $5$ are $0$ and $5$. $0$ can't be a leading digit, so the first and last digits must be $5$.

• Divisibility by $11$, every palindrome with an even number of terms is divisible by $11$. This follows from the alternating sum of the terms is $0$.

• Divisibility by $3$, $5+5=10$ which is two less than a multiple of $3$. Since we have two spots, we put $1$ and $1$ in the middle.

The only palindrome smaller than this with four digits, beginning and ending with $5$ is $5005$, which is not divisible by $3$.

What about palindromes with fewer digits? Well, if there are two digits, you get $55$, which isn't divisible by $3$. If you want three digits, divisibility by $3$ leaves only $525$, $555$, and $585$. None of these are divisible by $11$.

Assuming $0$ is cheating