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I would like to square a three digit number in my head. Now I know that the formula is

$$ ( X + r ) ( X - r ) + r^2 = X^2 - rX + rX - r^2 + r^2 = X^2 $$ Where $\,r\,$ is a number such that $\,X + r\,$ is divisible by $10$ and/or $100$

Now the problem is that, I would like to first figure out whether finding the $r$ such that the $3$ digit number is divisible by $100$ or $10$ would be more efficient.

If I choose $r < 10$ then the $\,( X + r ) ( X - r )\,$ becomes quite unpleasant. If I choose $r < 100$ then the $r$ itself will have to be broken into pieces such that $$( X + r ) ( X - r ) + [ ( r + j ) ( r - j ) + j^2 ] = ( X + r ) ( X - r ) + ( r^2 - jr + jr - j^2 + j^2 ) = $$ $$ =( X + r ) ( X - r ) + r^2 = X^2 - rX + rX - r^2 + r^2 = X^2$$

This kind of yields recursive solution where any number squared works. The job is much more easier, but the person has to memorize more numbers while continuing to do more math.

So which method would you recomend? Or if you would have some other method I would love to hear it.

If you choose recursive method how do you memorize numbers in an efficient way without having numbers collide in your brain?

Thanks to anyone for their response, and sorry if tags are off, I am unsure which tag fits my question.

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I edited a little your post but for some reason the middle rows I couldn't. Check this, please. –  DonAntonio Jun 12 '12 at 2:41
    
@DonAntonio Thank you very much for the edit. I have spent quite some time trying to figure out how to get the power to work. –  Quillion Jun 12 '12 at 2:56

2 Answers 2

up vote 3 down vote accepted

Many mental calculations are easier the more facts you memorize. Knowing the squares up to $31^2=961$ can make it easier, reducing the need to go all the way to single digits. Also the fact that $(10n+5)^2=100n(n+1)+25$, for example $65^2=(6\cdot7)25=4225$. If you break it all the way to individual digits, I find it easier to keep track of my place if I start with the most significant digits and add as I go, so $359^2=300^2+2\cdot300\cdot50+\ldots =90000+30000+\ldots$ $=120000+ 50^2+\ldots = 122500+2\cdot300\cdot9+\ldots=127900+2\cdot50\cdot9+9^2$ where for many purposes you can quit early when you have the precision needed.

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Thanks so much, I found this very easy and good to use. Now all that's left is to master it :) –  Quillion Jun 12 '12 at 13:26

Say is $$n = \left( {a,b,c} \right)$$

$$n=a \cdot 10^2+b \cdot 10+c$$

$${n^2} = {a^2}\cdot{10^4} + 2ab{10^3} + \left( {2ac + {b^2}} \right)\cdot{10^2} + 2bc10 + {c^2}$$

$${n^2} = \left( {{a^2},2ba,2ac+b^2,2cb,{c^2}} \right)$$

That is a nice symmetric formula in $a,b,c$ if you have to ask!

Note $ a^2 ,b^2, c^2 $ occur precisely in $1$st $3$rd $5$th, and you can always get the middle string right by thiking loopy:

$$baaccb$$

Example

$n=721 = (7,2,1)$

Then

$$n^2 = (49,2 \cdot 7 \cdot 2, 2 \cdot 7 \cdot 1+4, 2 \cdot 1 \cdot 2 , 1^2)$$

$$n^2 = (49,28, 18,4 , 1)$$

Then we do the "carries"

$$n^2 = (51,9, 8,4 , 1)=519841$$

I don't know if this the greatest method but it is interesting to consider, since at most you need to square one digit numbers.

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