This method is similar to squaring by Yaavadunam. It is just modified a bit, as we shall see in the next few examples.

Consider 133

Step 1 : Consider nearest base (here 10).

Step 2 : As 13 has a excess of '3' (13 - 10 = 3), we double the excess and add the original number (13) to it, and put it on the LHS.

Therefore we get 13 + 6 = 19

Step 3 : Now find the new excess. In this case it is 19-10 = 9. Now multiply this with the original excess to get the middle part of the answer.

Therefore we get 9 * 3 = 27

Step 4 : Now cube the original excess and put it as the last part

Carry over any big numbers and total to get the answer.

```
19  7  7
2  2
21  9  7
```

Therefore 133 = 2197

Now consider 473

As in 'Nikhilam' and Squaring, we use 'Aanurupyena' here.

1) Let the main base be 10 and the working base be 50

therefore the ratio

x = (Main Base)/(Working Base) = 10/50 = 1/5

2) Excess is -3 (47 - 50 = -3).  Double the excess and add the original number (here 47) to it.

We get 47 - 6 = 41.

The Base correction for this part is achieved by dividing by x2 .

therefore we get 41/(1/25) = 41 * 25 = 1025

3) Excess in the new uncorrected number (41 - 50 = -9) is multiplied by the original excess(-3) to obtain the second part.

Therefore we get -9 * -3 = 27

The Base correction for this part is achieved by dividing by x .

therefore we get 27 * 5 = 135

4) The third part is obtained by cubing the excess.

(-3)3 = -27

5) Carry over the extra numbers and total to obtain the final answer

```
1025  0  0
13  5  0
-2  7
1038  2  3
```

Therefore the final answer is 103823