Let 'abcd' represent the four digit number. For a four digit number, there is a 'thousands' column, a 'hundreds' column, a 'tens' column and a 'units' column. As the 'a' is in the 'thousands' column, its actual value it 1000a. The 'b' is in the 'hundreds' column, therefore its actual value is 100b. Similarly, the 'c' is in the 'tens' column and therefore its real value is 10c. As the 'd' is in the 'units' column, its value is just d. So, writing out 'abcd' algebraically, we get:
1000a+100b+10c+d
Now, as we add on the number 'bcda' (as we have to move the numbers from 'abcd' to the left and therefore the a comes to the 'units' column), we'll do the same as we did for 'abcd' for 'bcda':
1000b+100c+10d+a
We can now add the two algebraic expressions for the integers 'abcd' and 'bcda' by collecting like-terms and we'll get:
(1000a+100b+10c+d)+(1000b+100c+10d+a)
=1001a+1100b+110c+11d
We can now factor out '1001a+1100b+110c+11d' and we'll get:
11(91a+100b+10c+d)
∴The sum of the two integers 'abcd' and 'bcda' is a multiple of 11 always.
As for the three digit number case, let 'abc' denote the three digit number. So, the 'a' is in the 'hundreds' column, therefore its actual value it 100a. As for 'b', it's in the 'tens' column, so its value is 10b. The 'c' is in the 'units' column and it's value is just c. If we write 'abc' algebraically, we get:
100a+10b+c
Now writing an algebraic expression (as we similarly did in the previous example) for 'bca', we get:
100b+10c+a
We now add the two algebraic expressions together to get:
(100a+10b+c)+(100b+10c+a)
=101a+110b+11c
This, however, can't be factored out using 11 and therefore the answer of the sum of 'abc' and 'bca' is not a multiple of eleven.
I will now do the same as has been done above for the previous two examples, except for a five digit number. Let 'abcde' denote a five digit number. The a is in the 'ten thousands' column, so it actually represents 10000a. 'b' is in the 'thousands' column, therefore, it represents 1000b. As for 'c', it's in the 'hundreds' column, so its genuine value is 100c. The 'd' is in the 'tens column, so it's real value is 10d. As the 'e' is just in the 'units' column, so it's just an e. Writing 'abcde' algebraically, we get:
10000a+1000b+100c+10d+e
Now, writing out 'bcdea' algebraically, we will get:
10000b+1000c+100d+10e+a
Adding the two five-digit integers, we get:
(10000a+1000b+100c+10d+e)+(10000b+1000c+100d+10e+a)
=10001a+11000b+1100c+110d+11e
The sum of the integers 'abcde' and 'bcdea' can't be factored using 11 and therefore, the sum of the integers 'abcde' and 'bcdea' is not a multiple of eleven.
For the case of a six digit integer, we'll let 'abcdef' denote the integer. 'a' is in the 'hundred thousands' column, so it's 100000a. 'b' is in the 'ten thousands' column, so its a 10000b. The 'c' is in the 'thousands' column, so it's a 1000c. 'd' is in the 'hundreds' column, so it's a 100d. 'e' is in the 'tens' column, so it's a 10e. 'f' is in the 'units' column, so it's just an f. Writing the integer 'abcdef' algebraically, we get:
100000a+10000b+1000c+100d+10e+f
Writing the integer 'bcdefa' algebraically, we get:
100000b+10000c+1000d+100e+10f+a
Adding 'abcdef' and 'bcdefa' together, we get:
(100000a+10000b+1000c+100d+10e+f)+(100000b+10000c+1000d+100e+10f+a)
=100001a+110000b+11000c+1100d+110e+11f
This can be factorised to give us:
11(9091a+10000b+1000c+100d+10e+f)
So, the sum of 'abcdef' and 'bcdefa' is a multiple of eleven.
I did some other digit-numbers in my own time and I came to conclude that if you're doing the sum for numbers with an even amount of digits, then the answer will be a multiple of eleven. If the numbers have an odd number of digits, then the answer will not be divisible by 11.Therefore, if you used a 38-digit-number, the sum would be divisible by 11.