Subtraction Without Borrowing: Subtraction by Addition:

In our Math Tip last week, we showed one method of avoiding borrowing in subtraction. This is important not only to facilitate mental calculation but more importantly, it will minimize possible errors due to striking out or rewriting the figures in the minuend.

We will now present the second technique which we discussed in our book “25 Math Short Cuts” to avoid borrowing. We call this “Subtraction by Addition”.

The term may seem to be self-contradictory but it really means that we will simplify subtraction by using the 10’s complement of the subtrahend.

Let us take last week’s problem as an example: 12,432 – 9,789.

               12, 432 + 211 = 12, 643

  9,789   + 211 = 10, 000

               12,643 – 10,000 = 2, 643

We can see that by adding 211, which is the 10’s complement of the subtrahend to the minuend 12, 432 will give 12, 643, while 9,789 and 211 will complete a whole 10,000 when combined. The subtraction will then transform into an easy 12, 643 – 10,000 = 2, 643.

This can easily be proven algebraically.

Let m – minuend

       s – subtrahend

       b – base

       t – ten’s complement of the subtrahend

       d – difference            

we have  s = b – t and

d = m – s = m – (b – t)

   = m – b + t

  = m + t – b = (m + t) – b

Since we can easily determine the 10’s complement of a number using the sutra All from 9 and the Last from 10, the subtrahend can be far from the base, which is the case with this week’s example: 54,725 – 35, 947. This subtraction will require 4 regroupings or borrowing using the conventional method.

Now instead of performing that subtraction we added the 10’s complement of the 35, 947 which is 64, 053 to the minuend, to get the sum 118, 778 and we can deduct the base, 100, 000 by simply striking out the leading 1 in the answer.

This technique is particularly useful in computations like the following:

     + 89, 764                                                   89, 764

      -54, 367                                                      45, 633

      +76, 362                                                     76, 362

       -25,748                                                      74, 252

       -42,746                                                      57, 254                                  

                                                                          343, 265 – 300,000 = 43, 265

We can see that we replaced the numbers to be subtracted in the left group by their complements in the right group. So we can proceed to addition column by column in the right group. Since we replaced to three numbers by their complements, we will subtract 3 imes their bases or 3 x 100,000 = 300,000.

Try to solve the following exercises using subtraction by addition:

1. 342 – 289 =
2. 734 – 497 =
3. 4, 125 – 3, 879 =
4. 5, 245 – 4, 868 =
5. 8, 124 – 3, 766 =
6. 9, 232 – 5, 768 =
7. 73, 514 – 27, 647 =
8. 92, 465 – 38, 878 =
9. 86, 485 – 23, 776 – 43, 554 =
10.  88, 543 – 65, 374 + 43, 923 – 48, 778 =  

Here are the answers to last week’s practice exercises:

1. 222 – 198 = 24
2. 334 – 289 = 45
3. 715 – 476 = 239
4. 1, 123 – 987 = 136
5. 5, 323 – 889 = 4, 434
6. 6,145 – 798 = 5, 347
7. 23,456 – 9, 897 = 13, 559
8. 55, 432 – 49,678 = 5, 754
9. 53, 321 – 48, 787 = 4, 534
10. 1,222,111 – 777,889 = 444,222