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?
Best Answer
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.
$$x^2=(10a+b)^2=100a^2+10(2ab)+b^2$$
We can also flip the digits (I'll use $\bar x$ to indicate this) and then square.
$$\bar x^2=(10b+a)^2=100b^2+10(2ab)+a^2$$
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:
$$x^2=(100a+10b+c)^2=10000a^2+1000(2ab)+100(b^2+2ac)+10(2bc)+c^2$$
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.
$$100,101,102,103,110,111,112,113,120,121,122,130,200,201,202,210,211,212,220,300,301,310,311$$
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.