Question

Find the cube of the following numbers using Nikhilam Sutra
(a) $103$ (b) $105$ (c) $107$

Answer 1

By Nikhilam sutra, cube$=s^2(No.+2d)/s\times 3d^2/d^3$

where $s=$sub-base, $d=$difference

(i) $103$

base$=100$;

sub-base$=\left[103/100\right]=1$

$d=103-(100\times 1)=3$

Cube$=1^2(103+2(3\left)\right)/1\times 3\left(3^2\right)/3^3$

As base is $100$, we'll take two digits $109/27/27$

$1092727$

(ii) $105$

base$=100$; sub-base$=\left[105/100\right]=1$

$d=105-100\times 1=5$

Cube$=105+2\left(5\right)/3\left(5^2\right)/5^3$

$=115/75/125$

$=1157625$

(iii) $107$

base$=100$; sub-base$=\left[107/100\right]=1$

$d=107-(100\times 1)=7$

Cube$=107+2\left(7\right)/3\left(7^2\right)/7^3$

$=121/147/343$

$=1225043$.