Abacus Math Program – Lesson 11 – The 5/10 Pair Combination Part 1
So far in Lesson 5, 6, 7, and 8 we introduced addition and subtraction with 10 pair complements when the operation requires more than 1 rod. In Lesson 9 and 10 we introduced addition and subtraction with 5 pair complements when the operation is done on a single rod, i.e. no carry or borrow is required from another rod. Now here in Lesson 11 we will combine both the 5 and 10 pair complements in a single bead movement. In Lesson 11 Part 1 we will look at examples that need to perform a 10 pair complement addition but instead of subtracting the 10 pair directly we subtract in the form of a 5 pair complement. In all these cases we will need to subtract a 10 pair less than 5 but there will not be a sufficient number of lower earthly beads touching the bar to subtract the 10 pair complement directly. So we will need to use the upper heavenly 5 bead to complete the movement.
Let’s look at the example problem 7 + 7. To add the second 7 to the first we use the 10 pair addition rule. So first add 1 to the next rod left and then subtract 3, the 10 pair complement of 7, from 7 on the target rod. However to subtract 3 from 7 we need to use the 5 pair subtraction rule since we only have 2 lower earthly beads touching the reckoning bar. Therefore to subtract the 10 pair complement 3, in the form of a 5 pair we add 2 earthly beads, 3’s five pair complement, and subtract the 5 bead leaving the answer of 14.
1. 5+6, 5+7, 5+8, 5+9
2. 6+6, 6+7, 6+8
3. 7+6, 7+7
4. 8+6
In this first example we start by adding 17 to the abacus adding 1 to 0 on rod A and 7 to 0 on rod B. Next we add 6 to 17 by adding 6 to 7 on rod B. Notice this addition will require using the 5/10 pair combination. First doing the 10 pair addition we add 1 to 1 on rod A and then subtract 4, the 10 pair complement of 6, from 7 on rod B. But since there are only 2 lower beads touching the bar we will need to subtract the 4 as a 5 pair. So subtract the 4 by adding 1, the 5 pair complement of 4, and subtract the 5 bead. The interim sum is now 23. Next add 2 to 23 by adding 2 to 3 on rod B using the 5 pair addition. Add 5 and subtract 2, the 5 pair complement of 3, on rod B for an interim sum of 25. Lastly subtract 8 from 25 by subtracting 8 from 5 on rod B using the 10 pair subtraction rule. First subtract 1 from 2 on rod A and then add 2, the 10 pair complement of 8, to 5 on rod B. The final answer is 17.
In this next example we start by adding 5 to the abacus on rod B. Next we add 9 to 5 on rod B using the 10 pair addition rule. First add 1 to 0 on rod A and then subtract 1, the 10 pair complement of 9, from 5 on rod B. To subtract 1 from 5, we will need to use a 5 pair subtraction. So add 4, the 5 pair complement 1, to rod B and then subtract 5 for an interim sum of 14. Next add 8 to 14 by adding 8 to 4 on rod B. Again we will use a 10 pair addition by adding 1 to 1 on rod A and then subtract 2, the 10 pair complement of 8, from 4 on rod B for an interim sum of 22. Finally we add 3 to 22 by adding 3 to 2 using a 5 pair addition. So add 5 and then subtract 2, the 5 pair complement of 3 on rod B. The final answer is 25.
In the last example we start by adding 15 to the abacus adding 1 to 0 on rod A and 5 to 0 on rod B. Next subtract 7 from 15 by using the 10 pair subtraction rule. First subtract 1 from 1 on rod A and then add 3, the 10 pair complement of 7, to 5 on rod B for an interim sum of 8. Next subtract 2 from 8 on rod B for an interim sum of 6. Lastly add 8 to 6 on rod B using the 10 pair addition rule. So first add 1 to 0 on rod A then subtract the 2, the 10 pair complement of 8, from 6 on rod B. To complete the subtraction we will need to use a 5 pair subtraction. So subtract 2 from 6 by adding 3, the 5 pair of 2, and then subtract 5 for a final answer of 14.
