Abacus Math Program – Lesson 5 – Ten Pair Complement Addition

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In lesson 5, we will introduce the concept of 10 pair complements or 10 pairs. If you are new to abacus math, we suggest that you start from lesson 1. This abacus math lesson will use the 10 pair complement method for addition but as we will see in a later lesson the 10 pair complements are also used for subtraction. There are only 5 possible 10 pair complements: 9-1, 8-2, 7-3, 6-4, 5-5. You will notice each pair of numbers adds exactly to 10. This is not some strange coincidence but it is an exposure of the underlying fundamental math concepts of our base 10 number system. Having students understand not only do these 10 pair complements exist but how to use them in math calculations gives students a much deeper understanding of our number system and how it works. 

Because a single rod on the abacus or soroban can only represent a number between 0 and 9, we need a method for cases where adding a number to a rod results in a number higher than 9. In traditional methods of teaching math arithmetic, we would consider this a “carry” situation. But abacus and soroban math users think of the carry in terms of 10 pair complements, which is fast and efficient. As we have seen numbers greater than 9 require more than 1 rod. So the 10 pair complement addition method will help us manage arithmetic operations that involve more than 1 rod.

Just like your computer, the most efficient way to add numbers is using the 10 pair complements which is very easy to perform on the abacus. We need a simple test to know when we should use the 10 pair complements. If the number of available beads remaining on the rod, i.e. the number of beads NOT touching the reckoning bar, is less than the number we want to add on the target rod, we need to use complements. 

The complement addition rule is very simple. We first add 1 to the next rod left of the rod we are trying to add the number and SUBTRACT the 10 pair complement of the number we want to add from the target rod.

In the following examples, numbers within a circle show the order in which the beads should be moved. Numbers at the tip of an arrow show the value of the beads to be moved.

In this first example we start by adding 7 to rod B on the abacus. Next we add 5 to 7 on rod B. But notice that after placing the 7 on rod B we only have 2 remaining beads which is not enough to add the number 5. So this is where we need to use the 10 pair complement addition rule. The rule says, first add 1 to the next rod left on the abacus, rod A, using our thumb. Secondly the rule says, subtract 5, the 10 pair complement of 5. So we subtract 5 from 7 on rod B by moving the upper 5 bead away from the bar with our index finger. The interim sum on the abacus is now 12. Next we add 2 to 12 by adding 2 to 2 on rod B for a new interim sum of 14. Lastly we subtract 3 from 14 by subtracting 3 from 4 on rod B for a final answer of 11.

In this next example, we start by adding 2 to the abacus on rod B. Next we add 7 to 2 on rod B by pinching the upper 5 bead and 2 lower beads with our index finger and thumb. The interim sum on the abacus is now 9. Next we add 6 to 9 on rod B but notice there are no available beads to add 6 so again we need our 10 pair complement addition rule. So first we add 1 to the next rod left on the abacus, rod A. Then we subtract 4, the 10 pair complement of 6, from 9 on rod B resulting in the interim sum of 15. Lastly we subtract 5 from 15 by moving the upper 5 bead away from the bar on rod B for a final answer of 10.

In this last example, we start by adding 5 to the abacus on rod B. Next we add 4 to 5 on rod B by moving 4 lower beads to the bar with the thumb. The interim sum on the abacus is now 9. We then add 8 to 9 on the abacus but notice again there are no available beads on rod B so we need to use the 10 pair complement addition rule. So to add 8, we first add 1 to the next rod left on the abacus, rod A. Then we subtract 2, the 10 pair complement 8, from 9 on rod B. The resulting interim sum is now 17. Lastly we subtract 6 from 17 by subtracting 6 from 7 on rod B by moving the upper 5 bead and 1 lower bead away from the bar. The final answer is 11.

Next up Lesson 6 Ten Pair Complements Part 2.