For the past two weeks we have done Nikhilam multiplication with the base as 100. We now move on to bigger numbers and we can see how exciting it becomes.
Look at this big problem: 9999 x 9988.
We take 10000 as our base.
9999 is 1 less than 10000 and 9988 is 12 less than 10000
9999 - 12 = 9987
9988 - 1 = 9987
12 x 1 = 12
0012 (12 is preceded by 00 because the base has four zeros so the number of digits should be four)
The answer is 99870012
One more amazing example: 100006 x 100009.
Here the base is 100000
100006 is 6 more than 100000 and 100009 is 9 more than 100000
100006 + 9 = 100015
100009 + 6 = 100015
9 x 6 = 54
00054 (54 is preceded by 000 because the base has five zeros so the number of digits should be five)
The answer is 10001500054
Wasn’t that simple!
One thing to keep in mind is the number of zeros in the base and prefix zeros to the second part of the answer to make that many digits.
Here are some more examples
A. 9999987 x 9999954
9999987 - 33 = 9999954
9999967 - 13 = 9999954
13 x 33 = 429
0000429
The answer is 99999540000429
B. 99789 x 99997
99789 - 3 = 99786
99997 - 211 = 99786
3 x 211 = 633
00633
The answer is 9978600633
C. 10109 x 10003
10109 + 3 = 10112
10003 + 109 = 10112
3 x 109 = 327
0327
The answer is 101120327
Some problems for you to try:
10108 x 10002
10099 x 10003
9996 x 9995
999991 x 999990