Never Borrow or Carry the 1 Again
In grade school we learn to subtract multiple digit numbers going from right to left. We start by finding the difference at the ones place, then the tens place, then the hundreds, and so on. Going in this direction is easy if the digits of the top number exceed the digits of the bottom number in each place, since you never have to borrow a 1 and carry a 1 to make the difference positive. For example, 843 - 611 is easy to compute in your head. You just calculate 3 - 1 = 2, 4 - 1 = 3, and 8 - 6 = 2 to get the answer 232.
But things get messy when some of the digits in the top number are less than the digits in the bottom number. For most people, it's not so easy to mentally calculate 812 - 657 or 790 - 589 using the right-to-left method because you have to carry the 1 at least once in each problem.
For most problems in mathematics there is more than one way to find the answer, and arithmetic is no exception. If you sometimes struggle with the right-to-left method of subtraction, you might find it easier to go in the opposite direction, which eliminates borrowing and carrying. This article explains the left-to-right technique, using 3-digit numbers as examples. You can extend this method to figure differences between 2-digit and longer numbers.
The Left-to-Right Method
The first step of the method is to find the positive difference between the digits in each place. For example, if we want to compute 812 - 657, the positive differences of the hundreds, tens, and ones places are respectively 2, 4, and 5.
Next, add the appropriate number of zeros to the end of the three numbers you found above according to their places. Place two zeros at the end of the difference in the hundreds place, one zero at the end of the difference in the tens place, and no zeros at the end of the difference in the ones place. Continuing with the example you get 200, 40, and 5.
Now you need to note whether the top or bottom digit is larger in each place of the subtracted quantities. If the top digit is larger, leave the difference in that place alone. If the bottom digit larger, make the difference in that place negative.
In the example of 812 - 657, the hundreds place difference is "normal" since the 8 is larger than the 6, so we keep the difference 200 as it is. But in the tens place, the 1 on top is smaller than the 5 on the bottom, so our 40 becomes -40 instead. In the ones place, the 2 on top is smaller than the 7 on the bottom, so our 5 becomes -5.
Finally take the three numbers 200, -40, and -5 and add them together. Negative numbers are subtracted, so it's equivalent to calculating 200 - 40 - 5. Doing this in stages gives you (200 - 40) - 5 = 160 - 5 = 155. Thus, the difference between 812 and 657 is 155.
Compute 507 minus 384. First we take the positive differences of the digits in each place. 5 - 3 = 2, 8 - 0 = 8, and 7 - 4 = 3.
Next, we add zeros to these numbers according to their places. Thus, 2, 8, and 3 become 200, 80, and 3.
Now we note which differences resulted from the top being larger than the bottom or the bottom being larger than the top. Since the top 5 is larger than bottom 3 in the hundreds place, 200 stays the way it is. Since the bottom 8 is larger than the top 0 in the tens place, 80 becomes -80. And since the top 7 is larger than the bottom 4 in the ones place, 3 stays positive.
Lastly we compute 200 - 80 + 3:
(200 - 80) + 3 = 120 + 3 = 123.
Therefore 507 - 384 = 123.
Is This Easier Than the Traditional Method?
Many people find this method easier once they get the hang of it, but some people find it difficult to learn new arithmetic techniques if the old ways are ingrained.