Claim: there are no palindromic primes in base 10 (primes that read the same from back to front as front to back, e.g. 313) within the range [1000, 2000].
Proof: Assume that there is such a prime $ p $. Then by the definition of a palindromic prime, $ p $ must be in the form:
$$ 1000 + 100k + 10k + 1 $$
where $ k $ is any integer between 0 and 9, inclusive. One way to arrive at the conclusion is to combine the middle terms of the equation, and then factorise the common multiple 11:
$$ \begin{aligned} p &= 1000 + 110k + 1 \\ &= 1000 + 109k + (k + 1) \\ &= 999 + 108k + 2(k+1) \\ &= (999 - 1) + (108 - 1)k + 3(k+1) \\ &= (999 - 2) + (108 - 2)k + 4(k+1) \\ &\ldots \\ &= (999 - 108) + (108 + 2)(k+1) \\ &= 891 + 110(k+1) \\ &= 11(81 + 10(k+1)) \end{aligned} $$
Since $ p $ is divisible by 11 then it is by definition, not a prime and hence there is no palindromic prime within the range [1000, 2000].