In the previous article, we had learnt about the use of Abacus, representation of numbers on Japanese Abacus, clearing the Abacus, working from left to right and complementary numbers. In this article, which is the first article on Abacus Maths, we shall have introduction to the procedure of carrying out abacus sums involving addition of single digit and multiple digit numbers. In subsequent articles we shall learn about subtraction, multiplication and division. Here, Abacus addition means carrying out addition operation by use of Abacus.

We shall see how to carry out addition problems for the following cases:

• Case 1: Abacus addition of two single digit numbers whose sum is less than nine.
• Case 2: Abacus addition of two single digit numbers whose sum is more than nine.
• Case 3: Abacus Addition of two multiple digit numbers.

## Case 1: Abacus addition of two single digit numbers whose sum is less than or equal to nine.

Let us do Abacus addition of 4 and 3, whose sum is 7, which is less than or equal to 9.

• First, we shall set up the number 4 on Abacus by moving all the earth beads upwards towards the beam on the unit rod.
• Now, in order to add 3, which is equal to 5 – 2, we shall move the heaven bead (in top row) down to add 5 and then move 2 earth beads down to subtract 2.
• We now have heaven bead and 2 earth beads in set position giving us the result of the Abacus addition as 7.

## Case 2: Abacus addition of two single digit numbers whose sum is more than nine.

Let us do Abacus addition of 4 and 8, whose sum is 12, which is greater than 9.

While doing Abacus sums involving addition of two single digit number whose sum is greater than nine, it is required to subtract the compliment of second number from the first number and one bead is moved up on the rod which is on the immediate left to the rod where addition is being carried out.

• Set number 4 on the unit rod B.
• Find the complement of second number with respect to 10, which is equal to 10 – 8 = 2
• Now subtract 2 from 4 on the unit rod B by moving the last two earth beads downwards leaving only two earth beads in set position.
• Move one earth bead upwards on the rod A (which is rod on immediate left of rod B)
• Effectively the above operations can be explained as – We know that 4 + 8 =12, so if we replace 8 with 10 – 2, we get 4 + 10 – 2 = 12 or 4 – 2 + 10 = 12. In order to achieve 4 – 2, we carried out step 3 and to add 10 we carried out step 4 above. ## Case 3: Abacus Addition of two multiple digit numbers.

Let us try the abacus addition of 21 + 6.

• In order to carry out this abacus sums, we first set number 21 on the abacus frame by placing value of 2 on rod G and of 1 in rod H. Remember to enter this number from left to right with 1 being on unit rod. • In order to carry out addition from left to right, we are required to add 2 + 0 on rod G because number 6 can be considered to be equal to number 06.
• Next, we are required to add 1 + 6 on the unit rod. The sum of these two numbers is less than or equal to 9, so we shall use Case 1 method explained above, whereby to add 6, we need to set the heaven bead by moving it downwards and move one earth bead upwards.
• As can be seen, Rod G has value of 2 and rod H has a value of 7. Let us now add 15 to the result of abacus addition of the above abacus sums, which is 27 + 15.

• First, 23 shall carry out 2 + 1 on rod G, because we are required to work from left to right. We shall utilize method explained in Case 1, by moving one earth bead upward to place number 3 on the rod, because the sum of these two numbers is less than or equal to 9. At his stage the result is shown as 37, which is not the final answer but only an intermediate stage.
• Next, we are required to carry out addition of 7 and 5. Since the sum is greater than 9, method in Case 2 shall be followed. We find complement of 5 with respect to 10, which is 5 itself. Subtract 5 from 7, by moving heaven bead up and we get answer as 2. At this intermediate stage abacus shall show 32.
• Last step is to move one earth bead upwards on the rod G, which is rod on Immediate left to H.
• By this operation, we get our final answer as 42 as shown on the abacus below. Tags: No tags