In the number 16783, the groups will be 1,67 and 83. In grouping, start
from the right side of the number. If you are left with a single digit in the
left, treat that as a group. These are clearly illustrated in the model exercises.
II. FINDING THE SQUARE ROOT OF A PERFECT SQUARE :
A. Example : 324
a) Set the number 324 in the abacus. That is, 3 must be set in the sixth column
from the right, 2 in the fifth and 4 in the fourth column.
b) Now ask. What is the highest perfect square in 3. (You are grouping this
number as 3 and 24). It is 1. Set the 1 by skipping one column to the left of
- By saying 1×1, clear 1 from the 3. Now you are left with 224.
c) Double the first digit of the square root and place it in the left extreme of the
abacus. That means, you have to set 2 in the extreme left of the abacus.
d) Now consider the 22 of the remainder. For setting the second digit of the
square root, many ways are available.
- You ask : how many 2’s in 22? It is 11. But you should not set two digits
in the quotient (here square root) at a time. Hence consider the highest
single digit number. It is 9. Set the 9 to the right of the first digit of the
square root 1. You will get 19.
- Set the 9 also in the right side of the number 2 which is already set in the
left extreme of the abacus.
- Now multiply the 9 of the 19 with the 29 and subtract the value from 224.
Multiply 9×2 and subtract the value 18 from 22. You will get 4 as remainder
and altogether, you are left with 44 in the abacus. Multiply 9×9 = 81. 81
cannot be subtracted from 44. That means, your assumed quotient 9 is
too big.
- Reduce 1 and set the value as 8. Now reset the dividend part as 224.
Multiply 8×28 and subtract from the 224. You will be left with no
