Remarkable reversible numbers

Michael P. Greaney

Remarkable reversible numbers

Mental arithmetic

Some tricks to help you with your mental arithmetic when you're dealing with reversible numbers.

Reversible numbers, or more specifically pairs of reversible numbers, are whole numbers in which the digits of one number are the reverse of the digits in another number, for example, 2847 and 7482 form a reversible pair. Reversible pairs prove interesting because of the unexpected way in which they allow addition and subtraction to be carried out — a way that facilitates mental arithmetic. This is best illustrated with a couple of examples.


Take a reversible pair such as $13$ and $31.$ The addition of these two numbers can in this case be performed quite simply by taking the sum of the two digits and multiplying it by 11:

  \[  1+3 = 4 \mbox{ and } 4 \times 11 = 44, \mbox{ so } 13+31 = 44. \]    

Let’s try this again with the pair $82$ and $28:$

  \[  8+2 = 10 \mbox{ and } 10 \times 11 =110, \mbox{ so } 82+28 = 110. \]    


Subtraction is carried out in a similar fashion, except the difference between the two digits is taken and the result multiplied by 9. For example, in subtracting $13$ from $31$ we have

  \[  3-1 = 2 \mbox{ and } 2 \times 9 = 18, \mbox{ so } 31-13 = 18. \]    

Similarly, for $28$ and $82$ we have

  \[  2-8 = -6 \mbox{ and } (-6) \times 9 = -54, \mbox{ so } 28-82 = -54. \]    

It works like magic. But why?

The underlying theory

Any positive whole number $a$ which is less than $100$ is written with two digits $x,y$:

  \[ a=10x+y. \]    

Its reverse is therefore

  \[ b=10y+x. \]    

For example, if $a=82$ and $b=28,$ we have $x=8$ and $y=2:$

  \[ 82 = 10 \times 8 +2 \mbox{ and } 28 = 10 \times 2 + 8. \]    

The sum of $a$ and $b,$ then, is

  \[ a+b = (10x+y) + (10y+x) = 11x + 11y = 11(x+y). \]    

Similarly, their difference is

  \[ a-b = (10x+y) - (10y+x) = 9x + 9y = 9(x-y). \]    

The result also holds if $a$ is a number that ends in $0,$ such as $a=10.$ In this case, $b=01 = 1.$

Triple digit numbers

Does something similar work for triple digit numbers? We can use the same method as above to derive the corresponding equations. A positive whole number $a$ with the digits $xyz$ is equal to

  \[ a = 100x + 10y+z. \]    

Its reverse will therefore be

  \[ b = 100z+10y+x. \]    

For the sum we get

  \[ a+b = (100x + 10y+z) + (100z+10y+x) = 101(x+z) + 20 y. \]    

For the difference we get

  \[ a-b = (100x + 10y+z) -(100z+10y+x) = 99(x-z). \]    
Reversible numbers

The equation for the sum looks more complicated than for double digit numbers, but the equation for the difference preserves the same degree of simplicity. The trouble is that it doesn’t really make the mental arithmetic easier as it now involves multiplying by 99.

There is another neat trick, however: simply treat the first and last digits as though they formed a two digit number. First, find the difference using the equation for a two digit number, that is, $9(x-z),$ and then place a nine between the two digits (assume a leading zero if necessary).

As an example, consider $862$ and $268.$ As we saw above, the difference between the two digit numbers $82$ and $28$ is $54.$ Now drop a $9$ in the middle to give

  \[ 862 - 268 = 594. \]    

The last step can be justified by factorising the coefficient of the equation for the difference.

  \[ 99(x-z) = 11(9(x-z)). \]    

Since $9(x-z)$ is a multiple of $9,$ the sum of its digits will also be $9$ (that’s a fact you can easily verify). We now resort to a short-cut for multiplying by 11 that you might have heard of: split the two digits (in this case of $9(x-z)$) and place their sum between them. As we have just seen, that sum is $9,$ which justifies our trick above.

The general case

Can we extend these results to numbers with more digits? In the case of a four-digit reversible number $xyzw$ the equations become

  \[ a+b = 1001(x+w)+110(y+z) \]    


  \[ a-b = 999(x-w)+90(y-z). \]    

For a general $n$-digit number, written as $x_1x_2x_3...x_ n$ the equations are

  \[ a+b = 10^{n-1} \left(x_1 + x_ n\right) + 10^{n-2}\left(x_2+x_{n-1}\right) + 10^{n-3}\left(x_3+x_{n-2}\right) + ...+\left(x_ n+x_1\right) \]    


  \[ a-b = 10^{n-1} \left(x_1 - x_ n\right) + 10^{n-2}\left(x_2-x_{n-1}\right) + 10^{n-3}\left(x_3-x_{n-2}\right) + ...+\left(x_ n-x_1\right). \]    

This looks a lot more complicated. There are limits, then, as to how far we can go in finding the sum and difference of two reversible numbers this way. But that does not detract from its usefulness within those limits!

Further reading

Find out more about reversible numbers in Michael P. Greaney's Little book of reversible numbers.

About the author

Michael P. Greaney

Michael P. Greaney is a writer with particular interests in astronomy and mathematics. Apart from the Little book of reversible numbers, he has written articles for a number of astronomical magazines and was a contributing author to the book Observing and Measuring Visual Double Stars (Springer, 2012).

This question is for testing whether you are a human visitor and to prevent automated spam submissions.