Vedas and Mathematics
Sutra 2: "All from nine and the last from 10" comes handy when we wish to subtract a number such as 578 from a number having zeros equivalent or more than the number of number being subtracted e.g. 1,000. Each figure in 578 is subtracted from nine and the last figure is subtracted from 10, yielding 422. The interesting thing is that this problem can be solved from left to right the same way we write and read text.
This sutra can be extended to solve problems such as 60,000 minus 987. We simply reduce the number left after taking out the 0s to account for the number being subtracted (i.e 60 left after removing the 3 left 0s) by 1 and then apply the sutra, to get the answer 59,013.
The concept gives rise to Sub-Sutra 7: "Whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency". Let me explain this further:
For instance, in computing 92 we go through the following steps:
Let us take 10 as our base. To get 9 to the base 10 we have to subtract 1 from 10.
Since decrease 9 further by 1 to 8. This is the left side of our answer.
On the right hand side put the square of the deficiency, which is 12. Hence the answer is 81.
Similarly, 82-->(8-2)*10+ 22 --> 64, 72 = 49.
For numbers above 10,we take the surplus. For example:
112 = (11+1)*10 + 12 = 121
132 = (13+3)*10 + 32 = 169
and so on.
This is based on the fact that a2= (a + b)(a − b) + b2 where a is the number whose square is to be found and b is the deficit (or surplus) from nearest factor of 10.
398 X 11 =
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Sutras
This sutra can be extended to solve problems such as 60,000 minus 987. We simply reduce the number left after taking out the 0s to account for the number being subtracted (i.e 60 left after removing the 3 left 0s) by 1 and then apply the sutra, to get the answer 59,013.
The concept gives rise to Sub-Sutra 7: "Whatever the extent of its deficiency, lessen it still further to that very extent; and also set up the square of that deficiency". Let me explain this further:
For instance, in computing 92 we go through the following steps:
Let us take 10 as our base. To get 9 to the base 10 we have to subtract 1 from 10.
Since decrease 9 further by 1 to 8. This is the left side of our answer.
On the right hand side put the square of the deficiency, which is 12. Hence the answer is 81.
Similarly, 82-->(8-2)*10+ 22 --> 64, 72 = 49.
For numbers above 10,we take the surplus. For example:
112 = (11+1)*10 + 12 = 121
132 = (13+3)*10 + 32 = 169
and so on.
This is based on the fact that a2= (a + b)(a − b) + b2 where a is the number whose square is to be found and b is the deficit (or surplus) from nearest factor of 10.