What is the reverse of a two digit number?

Let us observe the sum of a 2-digit number and a number formed by reversing its digits.

23 + 32 = 55

63 + 36 = 99

43 + 34 = 77

21 + 12 = 33

26 + 62 = 88.

Notice that all the numbers are divisible by 11. Now, are all such numbers divisible by 11?

Theorem: The sum of 2-digit numbers whose digits are reversed is divisible by 11. 

Explanation and Proof

All 2-digit numbers with digits a and b can be written as 10a + b. For example,

34 = 10(2) + 4 (a = 3, b = 4)
89 = 10(8) + 9 (a = 8, b = 9)
57 = 10(5) + 7 (a = 5, b = 7).

In effect, when the digits are reversed, the number become 10b + a.

43 = 10(4) + 3
98 = 10(9) + 8
75 = 10(7) + 5.

So, in general, we have

10a + b: number
10b + a: number with reversed digits.

So, adding the two numbers we have

10a + b + 10b + a = 11a + 11b = 11(a + b).

As we can see, 11 is a factor of 11(a + b). Therefore, 11(a + b) is divisible by 11.

So, the sum of a 2-digit numbers and the same number with its digit reversed is divisible by 11.

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.

Addition

Take a reversible pair such as

and 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:

   

Let’s try this again with the pair

and

   

Subtraction

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

from we have

   

Similarly, for

and we have

   

It works like magic. But why?

The underlying theory

Any positive whole number

which is less than is written with two digits :

   

Its reverse is therefore

   

For example, if

and we have and

   

The sum of

and then, is

   

Similarly, their difference is

   

The result also holds if

is a number that ends in such as In this case,

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

with the digits is equal to

   

Its reverse will therefore be

   

For the sum we get

   

For the difference we get

   

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,

and then place a nine between the two digits (assume a leading zero if necessary).

As an example, consider

and As we saw above, the difference between the two digit numbers and is Now drop a in the middle to give

   

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

   

Since

is a multiple of the sum of its digits will also be (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 ) and place their sum between them. As we have just seen, that sum is 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

the equations become

   

and

   

For a general

-digit number, written as the equations are

   

and

   

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!

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Further reading

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

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About the author

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).

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Comments

Submitted by Anonymous (not verified) on 17 April, 2016

I've been trying to think of a method to decompose a number into a number which, added to its reverse, equals that first number. For example 685 decomposes into 392+293, but I only know that because I started with 392 arbitrarily. Do you or anyone know of a quick sure method to start with a number like 685 and find 392?

Chris G

• Reply

Submitted by Anonymous (not verified) on 23 April, 2016 In reply to by Anonymous (not verified)

Firstly, not all numbers have reversible components, e.g. 125.
Secondly, this solution applies only when the components have three digits, although the number itself could be as high as 1998 (= 999 + 999).
Let m be the number you want to decompose into its reversible components.
Let x and y be the components, then
x + y = m
The difference between two, three-digit numbers is evenly divisible by 99. So,
x – y = 99k
where k is some integer 0 … 9. It is the difference between the first and last digit of the larger component, as described in the above article.
Now,
y = m – x
x = 99k + y
On substituting m – x for y in the second equation and then adding x to both sides gives
2x = 99k + m
Or
x = (99k + m) / 2
The solution can be found by trying zero and even numbered values of k when m is even and odd numbered values m is odd.
There can be more than one solution to the problem. 685, for example has 3 solutions, namely when k = 1, 3 and 5. This means 685 has four reversible components: 392, 491, 590 and their respective reverses.

Michael Greaney

• Reply

Submitted by Anonymous (not verified) on 29 April, 2016 In reply to by Anonymous (not verified)

for your reply. However while it certainly added to my understanding of the topic, your method yielded no solution for any number that I pulled randomly out of the air, such as 341, 724, 651, 873 (like your example of 125). No doubt I'd have eventually hit on one, but it leads to the questions: what proportion of 3-digit numbers are decomposable this way, and is there a way of knowing in advance?

Interesting that my example 685 has three reversible components (not four as you say unless I've left something out), but applying your formula to it also yields 689 and 788 for k = 7 and 9. These are the two the 3-digit Lychrel (candidate) seed numbers, which I'm also interested in.

Also 651/2 = 325.5, which reversed and added a couple of times gives 651.156, which could be regarded as a fair approximation, would you say? Maybe that's the beginning of an alternative method.

When two numbers are reversed?

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.

What is the 2 digit number?

2-digit numbers are the numbers that have two digits and they start from the number 10 and end on the number 99. They cannot start from zero because in that case it will be considered as a single-digit number. The digit on the tens place can be any number from 1 to 9. For example, 45, 78, 12 are two-digit numbers.

What is the formula to reverse a number?

Reverse an Integer In each iteration of the loop, the remainder when n is divided by 10 is calculated and the value of n is reduced by 10 times. Inside the loop, the reversed number is computed using: reverse = reverse * 10 + remainder; Let us see how the while loop works when n = 2345 .

What is the reverse of 6?

The reverse of 9 is 6 and the reverse of 6 is 9. Step-by-step explanation: Hope it helps you...