Good solutions here came from Sam Larg, Dave Stewart, Richard Mason and Joe Neilson, Matthew Broadbent and Ross Craig, all in S3 at Madras College, St Andrews. Other people tested special cases but did not prove the general results.

If a 2 digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger we prove that this difference can never be prime.

Let the 2 digit number be $\text{'}\mathrm{ab}\text{'}$ where $a>b$. Then

${\displaystyle \text{'}\mathrm{ab}\text{'}-\text{'}\mathrm{ba}\text{'}=(10a+b)-(10b+a)=9(a-b).}$

As $9(a-b)$ is a multiple of 9, it is not prime.

Now let the 3 digit number be $\text{'}\mathrm{abc}\text{'}$

${\displaystyle \text{'}\mathrm{abc}\text{'}-\text{'}\mathrm{cba}\text{'}=(100a+10b+c)-(100c+10b+a)=99(a-c).}$

As $99(a-c)$ is a multiple of 99, it is not prime.

The 4 digit number can be taken as $\text{'}\mathrm{abcd}\text{'}$.

${\displaystyle \text{'}\mathrm{abcd}\text{'}-\text{'}\mathrm{dcba}\text{'}=(1000a+100b+10c+d)-(1000d+100c+10b+a)=9(111a+10b-10c-111d).}$

Again, for any 4 digit number, this difference is a multiple of 9 and so it can't be a prime number.

Similarly for 5 digit numbers:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\text{'}\mathrm{abcde}\text{'}-\text{'}\mathrm{edcba}\text{'}}& =(10,000a+1000b+100c+10d+e)-(10,000e+1000d+100c+10b+a)\\ \multicolumn{1}{c}{}& =99(101a+10b-10d-101e).\end{array}}$

This number is a multiple of 99 so it will never be prime.