Sequence of numbers with a special property

Question

Prove that the sequence a(n) = 2013 + 317n, where n is any nonnegative integer, generates infinitely many palindromic numbers.

Answer 1

Look for palindromic numbers of the most obvious format which also have a simple algebraic formula - the numbers $10^n+1$ are palindromic and are easy to deal with mathematically.

The equation $$10^n+1 = 2013 +317m$$

is equivalent to the congruence $$10^n \equiv 2012 \equiv 110 \pmod{317}$$

It is easy to see that this has solutions

$$n = 76, 155, 234, \dots$$

Thus the palindromic numbers $10^{76}+1, 10^{155}+1, 10^{234}+1, \dots $ all lie in your sequence, which hence contains infinitely many palindromic numbers.

For some hints for the mathematics behind this approach:

Since $\gcd(10,317)=1$ we know that $10^{316}\equiv 1 \pmod{317}$ (I "spotted" the fact that $10^{79}$ gives smaller numbers), so that we just have to find the first $n$ for which $10^n\equiv 110 \pmod{317}$, which is $n=76$.