Can you write the divisibility rule of 11?

Take any number of several digits Draw a vertical line somewhere in your scratch paper Write the left most digit in the left side of the vertical line Write the next digit in the right side of the vertical line Then the next digit again in the left side, and so on Now cancel out equal numbers from both sides Now cancel numbers that add up to same sum,  for example,  if we have 6 and 4 in left side and 8 and 2 in right,  we can see that, we're getting 10 in each case,  so,  cancel 6&4 with 8&2 We can cancel 6&2 with 8 also (for example) etc Continue Now,  we can split the Leftover numbers so that they are easier to cancel For example,  say,  we have 6, 4, 5 in the left side and 8, 3 in the right. We can re write the left side as 6, 2, 2, 5 and cancel 6&2 with 8. So, we're left with 2, 5 and 3. We'll rewrite as 2, 2, 3 and 3, so,  we can cancel 3 from both sides,  leaving no number (that is, zero) in right side and 2, 2 in the left Now,  add up the left and right sides separately and find the difference If that difference is divisible by 11, ONLY then the number will be divisible by 11 Please note,  zero is the universal multiple (that is,  divisible by ANY non zero number) And one is the universal divisor or factor Note,  you can start with the right most digit also. Also,  writing the first number in the left side of the vertical line is not mandatory,  you can write in right side also. Writing alternate numbers in alternate side is necessary This entire modification is my own discovery. Your constructive feedback needed. read less

If the difference between the sum of the digits in the odd numbered places( starting from the units place) and sum of the digits in the even numbered places is either 0 or multiple of 11, the number is divisible by 11. Ex 10824 sum of the digits in the odd numbered places - sum of the digits in the even numbered places (4+8+1) - (2+0) =11 ,divisible by 11. Hence, 10824 is divisible by 11. read less