Vedic Maths - Online Article
Vedic Mathematics
Multiplying by base method
This is applicable to multiplications where the numbers are close to a base like 10, 100, 1000 or so on. Let's take an example:
105 × 107. Here the base is 100 and the 'surplus' is 5 and 7 for the two numbers. The answer will be found in two parts — the right-hand should have only two digits (because base is 100) and will be the product of the surpluses. Thus, the right-hand part will be 5 × 7, i.e. 35. The left-hand part will be one multiplicand plus the surplus of the other multiplicand. The left part of the answer in this case will be 105 + 7 or for that matter 107 + 5 i.e. 112. The answer is 11235.
There can also be a carry-over from the right-hand part, e.g. 112 × 113. The right part will be 12 × 13, i.e. 156. But the right part should have only 2 digits. Thus, 1 will be carried over to the left part and the right part will be only 56. The left part will be 112 + 13 + 1 (the carry-over), i.e. 126. The answer will be 12656.
For 102 × 104 the answer will be 10608. Please note the right part will be 08 and not simply 8.
Can we use it for 92 × 97? Yes. In this case, the left part will be (–8) × (–3), i.e positive 24. The left part will be 92 + (–3) or 97 + (–8), i.e. 89. The answer is 8924.
How about 96 × 108? The right part will now be (–4) × 8, i.e. –32. To take care of the negative we will borrow 1 from the left part, which is equivalent to borrowing 100 (because we are borrowing from the hundred's digit of the answer). Thus, the right part will be 100 – 32 = 68. The left part will be 96 + 8 – 1 (the borrowed one) or 108 + (–4) – 1, i.e. 103. The answer will be 10368.
The same process can be done with other bases as well.
Corrolaries to base multiplication
Base multiplication can be used with other bases as well, like 20, 30 , 50 , 200, 500, etc.
24 × 28
Using base as 20 (which is twice of 10) do calculations as if you would do with base 10. Only thing to keep in mind is double the left-hand part (because 20 is twice of 10) before adding the carry-over if any. The right part will be product of surpluses from 10, i.e 4 × 8 = 32. But as per calculations for base 10, right-hand part will have only one digit. So 3 will be carried over. The left part will be twice of 24 + 8, i.e. 64 add to which the carry over 3. Thus, the answer is 672.
Squaring of numbers ending with 5
Multiply the number formed after ignoring the 5 in unit's place with the next integer to it. This is the left part of answer. Append a 25 as the right part, e.g. 652 is 6 × 7 with 25 appended to it, i.e. 4225; 1152 is 11 × 12 with 25 appended at the end, i.e. 13225.
Squaring of other numbers:
This method assumes that you are through with all squares of numbers up to 30.
A. If a number is in-between 30 to 80
Find out the 'surplus' (or for that matter the 'deficit') with 50 as base. The right part of the answer is the square of this surplus, just remember that the right part will have only 2 digits, all else will be carried over to left part. The left part is 25 + Surplus + Carry-over (if any)
572 is 25 + 7 // 72, i.e. 3249
382 is 25 + (–12) // (–12)2 i.e. 1444. In this case there is a carry-over of 1 from right part to left part.
B. If a number is in-between 70 to 130
The right-hand part of the answer is same as above with the surplus being calculated with respect to 100. The left-hand part of the answer is the given number itself + Surplus
1132 is 113 + 13 // 132 i.e. 12769.
742 is 74 + (–26) // (–26)2 i.e. 48 // 676, i.e 5476
Cubing of numbers close to 100, 1000, etc.
In this case, our answer will be in three parts. The right-most and the next part will have two digits each if base is 100 and three digits if base is 1000, as usual, the rest will be carried over if needed. The answer in parts will be: The number + Twice the surplus // Thrice the surplus square // Surplus cube.
Please note that surplus can also be a deficit, all we have to do is take care of the sign.
1053 is 105 + 10 // 3 × 25 // 53, i.e. 1157625
9983: This is a case with base 1000 and deficit 2. Thus, the answer is
998 + (–4) // 3 × (–2)2 // (–2)3
i.e. 994 // 012 // (–008)
i.e. 994011992
Multiplication
- When the sum of the digits of unit's place is 10
- When the rest of the digits are same
In this case, the right part of the answer is the product of unit's place (no restriction on number of digits). The left part is the remaining number (after eliminating the unit's digit) multiplied to the next integer, e.g. 13 × 17 is 1 × 2 // 3 × 7, i.e. 221. Another example 124 × 126... The answer is 12 × 13 // 4 × 6, i.e. 15624.
Multiplication when one multiplicand consists of only 9s
Let's do this with an example 3425 × 999. Since there are three 9s in the multiplicand, ignore three rightmost digits and take the remaining integer, in this case 3. Add 1 to this integer (always 1) to get 4. Subtract this 4 from the whole multiplicand, i.e. 3425 to get 3421. This is the left part of the answer.
To get the right part of the answer, take the complement of the three righmost digits of the multiplicand i.e. complement of 425 is 575. Thus, the answer is 3421575.
Another example 3236437 × 99999. There are five 9s. Ignoring five digits from the right we get 32, adding 1 we get 33, subtracting this from the whole multiplicand we get 3236404. The right part is complement of 36437, i.e. 63563. So the answer is 323640463563.
One more example, 456 × 999999. There are six 9s. So ignoring six digits from right we get 0, adding 1 we get 1, subtracting from multiplicand we get 455 as left part. Right part is the complement of 000456, i.e. 999544. So the answer is 455999544.
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