# Can someone explain these strange properties of 10,11,1210, 11, 12 and 1313?

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:

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),

As if this wasn’t enough, he also went on to mention that the same holds true for addition for those same four numbers!

So needless to say, this is hard to chalk up to mere coincidence. Is there some non-coincidental reason for these strange findings?

Suppose we have a 2 digit number $x$. We can write is in terms of its digits $a$ and $b$. When we attempt to square this number, we get an interesting result.

We can also flip the digits (I’ll use $\bar x$ to indicate this) and then square.

This result isn’t very useful on its own, but if $a^2$, $b^2$, and $2ab$ are all less than $10$, then the three terms above are the three digits of $x^2$ and $\bar x^2$ respectively. It is clear from that that reversing the digits of $x$ reverses the digits of $x^2$ provided it meets those requirements. (If we switch $a$ and $b$, the first and last terms switch while the middle term is unchanged.)

Note that $10$, $11$, $12$, and $13$ (as well as $20$, $21$, $22$, $30$, and $31$) all satisfy the same condition $a^2,b^2,2ab<10$ and thus have the property you describe.

The same argument can be used for addition, since, if no digit is greater than $4$, we can add the digits individually. Rearranging the digits of such a number will apply the same rearrangement to its sum with itself.

We can play the same game with three digit numbers, but the restrictions are even greater:

If each factor multiplying a power of ten is less than ten, we have the same property. This gives us a few numbers, which you can verify all have the property.

Note that we have not shown that these conditions give you all the numbers for which $\bar x^2=\bar{x^2}$, though I haven't been able to find a counterexample. This argument nonetheless applies to all of the numbers you provided.