TITLE: Reversing an integer's digits is multiplicative for small digits
QUESTION [91 upvotes]: So my 7 year old son pointed out to me something neat about the number 12: if you multiply it by itself, the result is the same as if you took 12 backwards multiplied by itself, then flipped the result backwards. In other words: $$12 × 12 = 144$$$$21 × 21 = 441$$
I confidently explained to him that this was merely a coincidence.
But he then casually pointed out that the same holds true for 10, 11 and 13 (as long as you use leading zeros), $$10 × 10 = 100$$$$01 × 01 = 001$$
As if this wasn't enough, he also went on to mention that the same holds true for addition for those same four numbers! $$12 + 12 = 24$$ $$21+21 = 42$$
So needless to say, this is hard to chalk up to mere coincidence. Is there some non-coincidental reason for these strange findings?

REPLY [10 votes]: It is not merely a coincidence. Rather, it is a special case of the fact that reversing the coefficients of a polynomial is a multiplicative operation (curiously this is not  well known, e.g. see prior answers).
Let $\,f = a_n x^n +\cdots a_1 x + a_0\,$ be a polynomial in $x.\,$ Reversing its coefficients yields
$\quad\ \ \bar f = a_0 x^n + \cdots a_{n-1}x + a_n = x^n f(x^{-1}),\ $ the reverse (or reciprocal) of $\,f.$
It is easy to show $\overline{fg}\, =\, \bar f\bar g,\,$ i.e. polynomial reversal is multiplicative. For example
$\qquad \begin{align} 
(1x+2)(1x+3)\, &= 1x^2+5x+6\, \overset{\large x\, =\, 10}\Longrightarrow\,  12\cdot 13\, =\, 156\\
\overset{\rm reverse}\Longrightarrow (2x+1)(3x+1)\, &= 6x^2+5x+1\ \ \Longrightarrow\,\ \ 21\cdot 31\, =\, 651
\end{align}$
Your examples are special cases when the product is a square (of linear polynomials). For the polynomials to yield integer reversals when evaluated at the radix $\,x=10\,$ it is necessary that all polynomials (including the product) have nonnegative coefficients less than the radix.