I have discovered a new way to obtain square of any number. I think its a new algorithm to find square of a number than just multiplying it with itself; if its not new then let me know. Algorithm is as follows:
- Divide the number in two parts with one part containing only the number at unit's place say part 'A', and other part say 'B', containing the remaining number.
- Now square the number at unit's place. The square will be one of these; {0,1,4,9,16,25,36,49,64,81}. The unit's place digit in this square is the unit's place digit in actual final answer.Write it in the answer. If the square of digit at unit's place is a two digit no like from 16 to 81 in above set; write only the digit at unit's place from this square in the final answer and carry the remaining digit.
- Multiply the actual number to be squared by part 'B'(the remaining part than the number at unit's place as described in step 1 ).
- Multiply the parts 'A' and 'B'.
- Add results of step 3 with results of step 4.
- Add the carried digit from step 2 to the sum in prior step, that is step 5.
- Now write this sum before the number we wrote at unit's place of final answer in step 2.
- This number we now obtain from step 7, is the square of our number.
Example:
Lets find square of 127 by above algorithm;
1. A = 7 and B = 12 here...
2. A^2 = 49 thus final answer will have 9 at units place and 4 is carried to add later.
3. Given number multiplied by 'B' => 127*12 = 1524
4. A*B => 12*7 = 84
5. 1524+84 = 1608 ..... ( step 3 + step 4 )
6. 1608+4 =1612 .... (4 is carried as stated in step 2)
7. Now, 16129 is the answer... (From step 2 and 6)
2 Answers
$\begingroup$Ok so let's try to square $10b+a$ and we let $a^2=10c+d$ where it is possible that $c=0$.
Step $1$ tells us to identify $a$ and $b$
For step $2$ we have $a^2=10c+d$ and we write $d$ in the final place and remember $c$
Step $3$ gives us $b\cdot (10b+a) =10b^2+ab$
Step $4$ gives us $ab$
Step $5$ gives us $10b^2+2ab$
Step $6$ gives us $10b^2+2ab+c$
In step $7$ "putting the number before" is the same as multiplying by $10$ and adding so we get $10\cdot(10b^2+2ab+c)+d$
To check at step 8 we have $100b^2+20ab+10c+d=100b^2+20ab+b^2=(10b+a)^2$
So your method works. I couldn't say whether it is new, but it requires two multiplications at steps $3$ and $4$, plus the squaring step.
There is an interestingly similar method for speeding up the multiplication of two large numbers. Let $M$ be a large power of $10$, say, so that the numbers $a, b, c, d$ are of comparable size.
Multiplying $(aM+b)(cM+d)=acM^2+(ad+bc)M+bd$ looks as though it requires the computation of four products $ac, ad, bc, bd$, but if we compute $ac, bd$ and $(a+b)(c+d)$ (three products) we find that $ad+bc=(a+b)(c+d)-ac-bd$. This turns out to be computationally more efficient.
The key thing about a method like yours is not so much whether it works (all sorts of methods work) or whether it is new, but whether it is a practical improvement on existing methods.
$\endgroup$ $\begingroup$$$\Large (10X+Y)^2=100X^2+20XY+Y^2$$
$\endgroup$ 3
