Tag: history of mathematics

12345  × 99 = 12345  × (100  - 1)
= 1234500 - 12345
= 1222155

9999  × 9 = 9999 (10 - 1) = 99990 - 9999

For 103^3, the base is 100 and deviation is +3. The formula is (number + 2*deviation) | (3*deviation^2) | (deviation^3), which gives (103 + 6) | (3*9) | 27 = 109 | 27 | 27 = 1092727.

The deviations from base 100 are -2, -3, and -1. The product is calculated by (98-3-1) | ((-2*-3) + (-3*-1) + (-1*-2)) | (-2*-3*-1), which is 94 | (6+3+2) | -6. Adjusting the -6 gives 94 | 11 | -6 = 94 | 10 | 94 = 941094.

The deviations from base 100 are -2, -1, and -1. The calculation is (98-1-1) | ((-2*-1) + (-1*-1) + (-1*-2)) | (-2*-1*-1), which is 96 | (2+1+2) | -2 = 96 | 5 | -2 = 964498.

The two rightmost digits in the multiplication 98,97, 99 = 100 - (2 + 3 + 1) = 100 - (2)(3)(1)

Using base 50 (sub-base 5*10), the deviations are +1, +2, +3. The calculation is 5^2 * (51+2+3) | 5*(1*2 + 2*3 + 3*1) | (1*2*3) = 25*56 | 5*11 | 6 = 1400 | 55 | 6 = 140556.

The two rightmost digits in the multiplication 98,97, 99 = 100 - (2 + 3 + 1) = 100 - 6 = 94

99  × 99 = 99 (100 - 1)
= 9900 - 99
= 9801

The correct representation of deviations from the base in the multiplication of 101,102,103 by Nikhilam formula of vedic maths is 01, 02 and 03.