Question

Vedic maths tricks to find square of any 4 digitnumber

Answer 1

For a number with four digits, say ‘abcd’, Duplex of ‘abcd’ => D(abcd) = 2ad + 2bc

The square of ‘abcd’ will have seven parts as shown below

abcd2 = D(a) | D(ab) | D(abc) | D(abcd)| D(bcd) | D(cd) | D(d)

= a2 | 2ab | 2ac + b2 | 2ad+2bc | 2bd + c2 | 2cd | d2

Example 5: Find the square of 1221

1221$^2$= D(1) | D(12) | D(122) | D(1221)| D(221) | D(21) | D(1)

= 1$^2$ | 2x1x2 | 2x1x2 + 2$^2$ | 2x1x1+2x2x2 | 2x2x1 + 2$^2$ | 2x2x1 | 1$^2$

= 1 | 4 | 8 | 10 | 8 | 4 | 1

As mentioned above only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.

Hence for 10 which has ‘1’ as the non units digit, we need to carry over ‘1’ to the immediate left which turns 8 to 9

1221$^2$ = 1 | 4 | 9 | 0 | 8 | 4 | 1

= 1490841