TITLE: When the product of a multi-digit integer and its mirror is a palindrome, can the original number have digits greater than $2$?
QUESTION [7 upvotes]: I am reposting a question I posted on r/mathematics. It was suggested I ask it here.
My son was jotting down some multiplications for school and asked me if there were many numbers that, when multiplied by their mirror image, resulted in a palindromic number (e.g. 221 x 122 = 26962).
I made a quick Python script to test this and found the results rather surprising.
For 3-digit numbers, there are 11 results. For 4-digit number, there are 23. The number of positive results doubles approximately with each addition of a digit, reaching 642 results with 9-digit numbers and 1118 results with 10-digit numbers. As you can see from the table below, the ratio of 2 seems to decrease with every iteration after the 6th.
This is the longest number we could test because calculating time increases exponentially by a factor of approximately 10, reaching 3 hours for the last example.

What I find interesting is that in all of the above results, with no exception, the factors are invariably composed of zeros, ones and twos. There's never anything else.
For example: 2100110011 x 1100110012 = 2310352049402530132.

I asked a mathematician friend — not remotely involved with number theory or arithmetic – and he said it might be related to "carry digits" messing things up. It's true that for 1-digit numbers, excluding the trivial zero, there are only 3 possible examples (1, 2 and 3) before the symmetry breaks (4 x 4 is 16 which isn't palindromic). But when multiplying huge 10-digit numbers you get tons of "carry digits" as can clearly be seen from the results: these can include any digit as seen in the example above.
It does seem to have some impact, though. For a test for n digits, all the multiplication results have the exact same number of digits, which is always 2n-1. e.g. 4-digit numbers always give 7-digit results.
I am sure there must be a deep reason for never seeing digits above 2 in the factors, but for the life of me I can't understand what it is.
Like I wrote I've only tested this up to ten digits, so my conclusion could be wrong.
Any insights are welcome. I'm not a mathematician, so please forgive me if this seems trivial to you.
I hope the table below is clear. Thanks a lot.
digits  digits  number       ratio       calc
in      in      of           with        time
factors results palindromes  previous
1       1       3       
2       3       4            1,333          0,001
3       5       11           2,750          0,001
4       7       23           2,091          0,011
5       9       46           2,000          0,110
6       11      93           2,022          1,081
7       13      185          1,989         10,973
8       15      353          1,908        108,295
9       17      642          1,819       1132,420
10      19      1118         1,741      11227,896

And BTW the script is below in case someone cares. I'm not a programmer either, so I wouldn't know how to multithread or otherwise optimize this, but it's a bit besides the point I think — the pattern here *does* seem to confirm itself, although of course it's no proof.
def mirror(length):
    print('Working...')
    count = 0
    start = time.time()
    for i in range(1, pow(10,length)):
        a = str(i).zfill(length)
        b = a[::-1]
        result = str(int(a) * int(b))
        if (result == result[::-1]):
            print(a, b, result)
            count += 1
    end = time.time()
    print(f'-----------\nSize : {length}\nTotal  : {count}\nTime  : {round(end-start, 3)}')
    
mirror(6)

REPLY [2 votes]: What a question!
If we don't worry about carrying:
$(10^3 a + 10^2 b+ 10c + d )= (10^3d + 10^2c + 10b + a) =$
$10^6ad + 10^5(ac+bd) + 10^4(ab+bc+cd) + 10^3(d^2+c^2+b^2+a^2) +10^2(ab+bc + cd) + 10(ac + bd) + ad$
And it's not hard to carry that out that if  
$N= \sum_{k=0}^n a_k10^k$ and $rev(N) = \sum_{k=0}a_{n-k}10^k$ and
$N*rev(N)= \sum_{k=0}^{2n} 10^k*(\sum_{j,i: i+j=k}a_ja_{n-i})$
And as: if $i+j =k$ then $(n-i)+(n-j) = 2n-k$ and $a_j= a_{n-(n-j)}$ and $a_{n-(n-(n-i))}=a_{n-(n-i)}$ so the coefficient for the $k$th digit in the product is the coefficient for the $2n-k$th term and the product is palindromic.
So all such products should be palindromic.  But if any $(\sum_{j,i: i+j=k}a_ja_{n-i})\ge 10$ we will have to carry and that screws everything up.
So two questions arise:
1) If any digit is more than $2$ will that force us to carry?
2) If we carry does that mean it cant be palindromic?
... 
1) Look at the $n$th term  $10^n(\sum_{i+j=n} a_ja_{n-i})= 10^n(\sum_{i+j=n}a_ja_j)=10^n(\sum{k=0}^n a_k^2)$ and so if any $a_j \ge 3$ then $a_j^2 \ge 9$ and if there is more that one significant digit... we have to carry.
[Note: Theoretically  $30000*00003 = 30000*3 = 90000 = 000090000$ is palindromic if we allow leading zeros but... that's garbage....]
2)  Yes.  If the first case where we have to carry is in the $k$th possition that means the $2n-k$th position will have to carry as well.  But the means we carry to the $2n-k+1$th position.  But the $k-1$th position wasn't carried to.  SO it can't be palindromic if we carry.
S0.....
Any number with a digit more than $2$ will force us to carry
And carrying always prevents palindromity
So the only palindromic results will not have any digit of more then $2$.
....
Oh, further more.  There can only be at most two $2$s and one $1$, or one $2$ and five $1$s,  or zero $2$s and nine $1$s.
....
And in base $b$ all digits must be strictly less than $\sqrt {b-1}$.