EKĀDHIKENA PŪRVEŅA
The Sutra (formula) Ekādhikena Pūrvena means:
“By one more than the previous one”.
Now let us apply this sutra to the ‘squaring of numbers ending in 5’.
Consider the example 25^2.
Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure 'to multiply the previous digit 2 by one more than itself, that is, by 3. It becomes the L.H.S (left hand side) of the result, that is,
2 X 3 = 6. The R.H.S (right hand side) of the result is 5^2, that is, 25.
Thus 25^2 = 2 X 3 / 25 = 6/25=625.
In the same way,
35^2= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;
65^2= 6 X 7 / 25 = 4225;
105^2= 10 X 11/25 = 11025;
135^2= 13 X 14/25 = 18225;
Now try to find out the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.
“By one more than the previous one”.
Now let us apply this sutra to the ‘squaring of numbers ending in 5’.
Consider the example 25^2.
Here the number is 25. We have to find out the square of the number. For the number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more than the previous one', that is, 2+1=3. The Sutra, in this context, gives the procedure 'to multiply the previous digit 2 by one more than itself, that is, by 3. It becomes the L.H.S (left hand side) of the result, that is,
2 X 3 = 6. The R.H.S (right hand side) of the result is 5^2, that is, 25.
Thus 25^2 = 2 X 3 / 25 = 6/25=625.
In the same way,
35^2= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;
65^2= 6 X 7 / 25 = 4225;
105^2= 10 X 11/25 = 11025;
135^2= 13 X 14/25 = 18225;
Now try to find out the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.
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