# Can someone explain these strange properties of $10, 11, 12$ and $13$?

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?

#### Solutions Collecting From Web of "Can someone explain these strange properties of $10, 11, 12$ and $13$?"

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.

Let’s start with addition. In a positional system like the one we ordinarily use, when we add two numbers, digits of different weights interact through carries. If there are no carries out of any positions, the sum may be performed on each pair of digits independently and then the overall result assembled from those partial results by just arranging them in the right order. For $0 \leq i < 5$, we have $2i < 10$; hence if all digits of a number are less than $5$ the addition trick will work.

The multiplication trick is similar. For instance,
$$112 \cdot 112 = 11200 + 01120 + 00224 = 12544 \enspace,$$
while
$$211 \cdot 211 = 42200 + 02110 + 00211 = 44521 \enspace.$$
Note that there are no carries generated, either while computing the partial products or while computing the final sum.

While the condition $0 \leq i < 5$ works for any numbers of digits when computing $n+n$, the increase in the number of partial products means that the multiplication trick only works for relatively small numbers. For instance,

$$11111112 \cdot 11111112 = 123456809876544 \enspace,$$

but

$$21111111 \cdot 21111111 = 445679007654321 \enspace.$$

The effect of the carries is initially visible in the middle digits, which depend on the most partial products.

I’ll make it really simple – you could explain to your son the exact reason after reading the answer.

Let the 2-digit number in question be represented as: $10a+b$

Now, $$(10a+b)^2=100a^2+b^2+20ab$$
On reversing its digit, you get: $$(10b+a)^2=100b^2+a^2+20ba$$

Notice that, a 3 digit number can be represented as $100x+10y+z$, and that in the above case, IFF $$20ab<100;b^2<10;a^2<10$$are satisfied, we can have numbers satisfying the property your son noticed. This is because, if any one of the 3 conditions above were not true, it would result in carried-over addition. Eg: For 14, $4^2<10$ is clearly not true. It can be seen that: $$14*14=196=100*1^2+4^2+20*1*4$$
And that:
$$41*41=100*4^2+1^2+20*4*1\ne196$$
So the only numbers satisfying it are: $00,01,02,03,10,11,12,13,20,21,22,30,31$

For the second part of your question, it can be seen that:
$$(10a+b)+(10a+b)=2(10a+b)=20a+2b$$
And on reversing the digits,:
$$(10b+a)+(10b+a)=2(10b+a)=20b+2a$$
Clearly, iff $20a,20b<100$ and $2a,2b<10$, i.e. $a,b<5$, the observation your son made holds.

So 14, unlike in the multiplication scenario, actually fulfills the above condition. (Hence $14+14=28$ & $41+41=82$)

So the only numbers satisfying it are: $00,01,02,03,04,10,11,12,13,14,20,21,22,30,31,33,40,41,44$

Readers may be interested to learn that this property is a special case of a property of polynomials. Namely: reversing the coefficients of a polynomial is a multiplicative operation.

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} (x+2)\ (x+3)\, &=\ \ x^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.

Motivated by the proofs above, I thought of the more general case:

Suppose that $a,b,c,d,x,y,z$ are integers such that $0 \le a,b,c,d,x,y,z < 10$. Furthermore, let $(10 a + b) \times (10c + d)= xyz$. Then, for what values of $a,b,c,d$ do we have $(10b+a) \times (10d + c) = zyx$?

For example, $14 \times 12 = 168$ and $41 \times 21 = 861$. Extending the proofs provided by others in the above posts to the present case, we have
$$\left(10 a + b \right) \times \left( 10c + d\right) = 100 ac + 10(ad + bc) + bd.$$

Similarly,
$$\left(10 b + a \right) \times \left( 10d + c\right) = 100 bd + 10(bc + ad) + ac.$$

Suppose $0 \le ac,bd < 10$ and $ad+bc <10$. Then, there is no carry-over and we have that $(10b+a) \times (10d + c) = zyx$.

This opens up a lot more possibilities. Fascinating stuff!