Abacus Math Program – Lesson 16 – Reverse Double Rod
In Lesson 11 we looked at the Forward Double Rod movement for addition problems. Now let’s take a look at the Reverse Double Rod movement for subtraction. The double rod movement for subtraction is exactly the same as for addition with the only difference using the 10 pair subtraction rule instead. Again we will repeatedly use the 10 pair subtraction to complete the more complex problems.
Let’s consider the problem 103 – 9. In this example we need to subtract 9 from 3 using the subtraction rule by first subtracting 1 from the next rod left. However the next rod left is zero so we must execute another 10 pair subtraction by subtracting 1 from 1 on the third rod and adding 9 to the second rod, the 10 pair complement of 1. We complete the first 10 pair by adding 1, the 10 pair complement of 9, to 3 on the first rod for an answer of 94.
In the first example we start by adding 97 to the abacus. Next we add 58 to 97 by first adding 5 to 9 on rod B using a 10 pair addition by adding 1 to 0 on rod A and subtracting 5, the 10 pair complement of 5, from 9 on rod B. We add 8 to 7 on rod C using a 10 pair addition by adding 1 to 4 on rod B using a 5 pair adding 5 and subtracting 4 leaving 5 on rod B. Next we subtract 2, the 10 pair complement of 8, from 7 on rod C for an interim sum of 155. Next we subtract 57 from 155 by first subtracting 5 from 5 on rod B leaving 0 on rod B. Then we subtract 7 from 5 on rod C using a 10 pair subtraction. First subtract 1 from 0 on rod B by doing another 10 pair subtraction and subtracting 1 from 1 on rod A and adding 9, the 10 pair complement of 1, to 0 on rod B. Finish the movement by adding 3, the 10 pair complement of 7, to 5 on rod C for a final answer of 98.
In the last example we start by adding 47 to the abacus. Next we add 148 to 47 by first adding 1 to 0 on rod C. Then adding 4 to 4 on rod B using a 5 pair by adding 5 and subtracting 1 leaving 8 on rod B. Next we add 8 to 7 on rod C using a 10 pair addition by adding 1 to 8 on rod B and then subtracting 2, the 10 pair of 8, from 7 on rod C for an interim sum of 195. Lastly we subtract 98 from 195 by first subtracting 9 from 9 on rod B leaving 0 on rod B. Next we subtract 8 from 5 on rod C using a 10 pair subtraction by subtracting 1 from 0 on rod B. To subtract 1 from 0 on rod B we use another 10 pair subtraction by subtracting 1 from 1 on rod A and adding 9, the 10 pair complement of 1, to 0 on rod B. Finally we add 2, the 10 pair complement of 8, to 5 on rod C for a final answer of 97.
