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Since we are adding 1 to the tens digit and subtracting one from the units digit, the sum of the digits should remain the same. For example, 9 + 2 = 11 with 1 + 1 = 2. When adding 9 to itself, we would thus expect the sum of the digits to be 9 as follows: 9 + 9 = 18, (1 + 8 = 9) and 9 + 9 + 9 = 27, (2 + 7 = 9).
Example of addition with carry. The black numbers are the addends, the green number is the carry, and the blue number is the sum. In the rightmost digit, the addition of 9 and 7 is 16, carrying 1 into the next pair of the digit to the left, making its addition 1 + 5 + 2 = 8. Therefore, 59 + 27 = 86.
During the addition, each carry is "signaled" rather than performed, and during the carry cycle, the machine increments the digits above the "triggered" digits. This operation has to be performed sequentially, starting with the ones digit, then the tens, the hundreds, and so on, since adding the carry can generate a new carry in the next digit.
Both these methods break up the subtraction as a process of one digit subtractions by place value. Starting with a least significant digit, a subtraction of the subtrahend: s j s j−1... s 1. from the minuend m k m k−1... m 1, where each s i and m i is a digit, proceeds by writing down m 1 − s 1, m 2 − s 2, and so forth, as long as s i ...
Subtraction is itself a sort of inverse to addition, in that adding x and subtracting x are inverse functions. Given a set with an addition operation, one cannot always define a corresponding subtraction operation on that set; the set of natural numbers is a simple example.
The sum of two biggest two-digit-numbers is 99+99=198. So O=1 and there is a carry in column 3. Since column 1 is on the right of all other columns, it is impossible for it to have a carry. Therefore 1+1=T, and T=2. As column 1 had been calculated in the last step, it is known that there isn't a carry in column 2.
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