This method is based on the digit position of each digit. You can think of each position as either an even or odd position. I am not talking about the digit itself, just the position. So the positions are 1, 2, 3, 4, 5, 6, etc. Which are 1, 3, 5, etc. as odd positions, while 2, 4, 6, etc. are the even positions.
If you add up all the even position digits and also add up all the odd position digits, you have two totals. Subtract one total from the other. This final result is a much smaller number. If this new number is divisible by eleven, then the original number was divisible by eleven. If your smaller number is still too large to tell if it is divisible by 11, then use the method again and make a still smaller number to test for divisibility by 11.
An example: Is 123456789 divisible by 11?
- 25-20=5, 5 is not divisible by 11, so 123456789 is not divisible by 11.
Next example: Is 1234567895 divisible by 11?
- 25-25=0, 0 is divisible by 11, so 1234567895 is divisible by 11 as well.
Ok, we have a method, but it has a weakness that it really needs pen and paper to do. But a simple adjustment in how we think about it, changes this into a method you can do in your head.
If you just walk along the digits in order, and add the first subtract the second, add the third, subtract the fourth digit, etc. You are performing the equivalent arithmetic of the above algorithm.
For example: Is 123456789 divisible by 11?
- 1-2+3-4+5-6+7-8+9= Wait, let’s do this in our head one sum at a time…
- 1-2=-1, -1+3=2, 2-4=-2, -2+5=3, 3-6=-3, -3+7=4, 4-8=-4, -4+9=5
- If you notice 5 is the same result we got in the first example. And 5 is not divisible by 11, so 123456789 is not divisible by 11 either.
For some numbers, there is a further short cut. Any time you see the same digit next to each other in pairs, you can skip them both. Because, of course, a number minus itself is zero.
For example: Is 322334455667788993 divisible by 11?
- First cross out all the pairs, and you are left with 3…………….3;
- But you can remove those positions and push the two digits together: 33
- Is 33 divisible by 11? You probably know already, yes. But notice, you can repeat the method and 3-3=0. 0 is always divisible by 11, or any number.
- So 322334455667788993 is divisible by 11.
So why does this work?
- A number ABCDEF is 100000A + 10000B + 1000C + 100D + 10E + F
- rearranging, this equals 100000A + 1000C +10E + 10000B + 100D + F
- 100000 divided by 11 is 9090 and remainder 10, or another way to look at it, it is 9091 and remainder -1
- If I divide 100000 by any factor of 100, or multiply by any factor of 100, the remainder is always the same, remainder = -1
- 100000A divided by 11 is (-1)A or a remainder of -A
- Similarly 10000 divided by 11 is 909 and remainder 1
- If I divide 10000 by any factor of 100, or multiply by any factor of 100, the remainder is always the same, remainder = 1
- 10000B divided by 11 is (1)B or a remainder of +B
- If you add several numbers together and divide by 11, you get the same remainder as you would get if you divide each of the several numbers by 11 and add up the remainders
- (100000A + 1000C + 10E + 10000B + 100D + F) / 11 is equal to 100000A/11 + 1000C/11 +10E/11 +10000B/11 +100D/11 + F/11
- The remainders for each part of the sum are (-A) + (-C) + (-E) + (+B) + (+D) + (+F)
- Rearranging, B + D + F – A – C – E equals the remainder for ABCDEF/11
- This will work for any number of digits
A more formal explanation using modular arithmetic might also help you understand.