**For my first Instructable**, I posted my first Instructable about a fun trick I had learned back in junior high for taking the cube roots of large numbers (6 or more digits) mentally. It was a well liked and well received 'ible, getting featured and making the popularity section in spite of being of being about math, so thank you very much for the warm welcome everyone!

The Instructable brought about many requests for additional mental math skills, if I happened to know any, and concerns about the practicality of mental cube root extraction. Of the comments, however, one in particular intrigued me. A lone user suggested I apply the concepts of Vedic Mathematics to my mental math repertoire. Having never heard of vedic mathematics, I went and found what I could off the internet. What I found was fascinating, and I will share with you a trick from Vedic Math today. So today, I will teach you how to square 2 digit numbers mentally, enjoy!

For more information on Vedic mathematics, including the background and history of Vedic Math, check this out.

**Signing Up**

## Step 1: A Note on Practice

So, if you find you can't solve mental squares as quickly as you'd like, one of the best ways to improve is to write out a whole bunch of multiplication problems on a sheet of paper, things like 25x46, 17x64, 7x920, and so on. Then, mentally solve them (don't write anything down, no calculators!) and finally, check your answers with a calculator. The more mental math problems you do, the better you'll be at it.

It should be noted that 2-digit integers will be used for the most part in this Instructable. However, the techniques presented here can be applied to

*decimals and numbers of arbitrary length*. With practice, it is possible to mentally calculate squares of decimals and 3 or more digit numbers. The only limit is your ability to remember numbers and multiply numbers mentally quickly.

## Step 2: Strategies for Computing Squares

Turns out there are a couple different ways:

1) Memorize every 2-digit number, and simply recite the square from memory (rote memorization).

2) Multiply the number by itself mentally (45*45)

3) Split 45 into components and apply the identity (a+b)

^{2}= a

^{2}+2ab+b

^{2}

4) Apply a high speed Vedic Mathematics technique for squaring numbers

As can be seen from above, the first method isn't general or practical. Even if you knew the squares of every 2 digit number, that's only 99 numbers (I'm including 1-9 here, I'm aware they are not 2-digits), and in the broader realm of math, won't get you very far.

Method 2 seems like a reasonable approach, though. However, the numbers you have to hold in memory are not very nice. At one time, you must remember that 5*45 is 225 and 40*45 is 1800 and add them in your head. For many numbers, this proves to be quite difficult.

Method 3 is a decent method, decidedly better than the first 2. We could express 45

^{2}in the following way:

45

^{2}= (40+5)

^{2}=40

^{2}+2*40*5+5

^{2}= 1600 + 400 + 25 = 2025

But, better than any of those is the 4th method, which I will show you next.

## Step 3: Mental Squares: The Vedic Way

"Whatever the extent of the deficiency, lessen it still further by that extent, and add to it the square of the deficiency"

That's a little cryptic, so below is the technique expressed in plain English:

For a given number, choose an arbitrary base. Now, determine the difference between the number and the base (this is the "deficiency"). Add the difference to the number, and multiply this new number by the base. Finally add the square of the difference to the number, which gives you the final answer.

That can be a little hard to digest, so we will apply the technique using the example of 45

^{2}from step 2.

1) Given 45

^{2}, choose 50 to be the base. Now, the deficiency is 45 - 50 = -5.

2) Add the deficiency to the number, so 45 + (-5) = 40.

3) Multiply the new number by the base: 40 * 50 = 4*5*100 = 2000

4) Add to the product the square of the deficiency: 2000 + (-5)

^{2}= 2000 + 25 = 2025

This technique, when practiced, allows for rapid mental square computation. More examples in the following steps.

## Step 4: Prove It!

So, we start with the following algebraic expression, which always holds true (foil out the right side and this can be seen):

x

^{2}-a

^{2}= (x+a)(x-a)

Now, simply isolate the x

^{2}term:

x

^{2}= (x+a)(x-a) + a

^{2}

The above expression is the mathematical form of the vedic sutra.

## Step 5: A Few More Examples

Example 1: Compute 67

^{2}

Choose a base (70).

Now, determine the deficiency (67 - 70 = -3), and add it to the number (67 + -3 = 64).

Multiply the base by the sum (64*70) = 7*64*10 = 448*10 = 4480

Finally, add the square of the deficiency to the product: 4480 +9 = 4489

And voila! There you go :-)

Example 2: Compute 12.5

^{2}

Choose a base (10)

Determine the deficiency (12.5 - 10 = 2.5) and add it to the number (12.5 + 2.5 = 15)

Multiply the base and the sum (15 * 10 = 150)

Add the square of the deficiency to the product: 150 + 2.5

^{2}= 150 + 6.25 = 156.25

Example 2 illustrates 2 important points. First, the method works even for non-integers, and second, the deficiency need not be negative.

In looking at the method, some of you may be wondering if the choice of base matters. Not at all. Suppose we redid example 1 using a base of 60 instead of 70:

Example 3: Compute 67

^{2}(using 60 as the base instead of 70)

Choose a base (60).

Compute the deficiency (67 - 60 = 7) and add it to the number (67 + 7 = 74)

Multiply the base and the sum (60 * 74 = 6*74*10 = 444*10 = 4440)

Add the square of the deficiency to the product: 4440 + 7

^{2}= 4440 + 49 = 4489

Your choice of base need not be a multiple of 10 either, but using multiples of 10 generally simplifies the math, so they are used most frequently.

whenever you have a number which ends in 5 you can use this. multiply the number preceding 5 with its successor. say you have 75 then multiply the preceding number 7 with it's successor 8 ( for 85 it would be 8*9), Write down the result and just write 25 to the end of this result.

So 75*75 = 7*8 = 56 now write 25 to the end of this digit = 56 25 this is the ans 5625.

The idea is to treat the number as 2 distinct numbers 1st part is the number preceding 5 and the second part is the digit 5.

so for 125 * 125 we get 12 & 5

12*13 = 156 and write 25 to the end = 156 25 ---> 15625

Method for checking.1. we get the digit sum of a no. by "adding across" the no. For instance, the digit sum 0f 13022 is 8.

2. we always reduce the digit sum to a single figure if it is not already a single fig.

3.In "adding across" a no. we may drop out 9's. Thus if we happen to notice two digits that add up to 9, such as 2 and 9, if we ignore both of them; so the digit sum of 99019 = 1 at a glance.(If we add up 9,s we get the same result)

4. because :nine don't count" in the process, as we saw in3 step, a digit sun of 9 is the same as a digit sum of zero. The digit sum of 441,e.g = 0.

You may use this forMultiplication, Cubes , Squares etc.you may also check weather your squared no, is correct or not.

I will explain it:-

take an example for 207.

square(207)= 42849

now add digits of LHS and RHS separately. we get

Sq(2+7) = 18

Sq(0) = 18

0=2+7

0=9

0=0

thus your calculation is correct.

take another example.

Sq(897) = 804609

sq(6) = 18

36 = 180

0=0.

Thanks,

Purduecer

Ever wondered why, for instance, 9*11 is one less than 10*10? 4*6 is one less than 5*5? 164*166 is one less than 165*165? (It is, trust me).

This is a specific instance of the identity

(x+a)(x-a) == x, called^{2}- a^{2}completing the square, and the Vedic squares method is another application of this. If you add a^{2}to each side you getx^{2}== (x+a) * (x-a) + a^{2}.For example, if x=67 and a=3,

x * x = 67 * 67

(x+a)*(x-a) + a*a = (70 * 64) + 3*3

= 4480 + 9

= 4489

= 67

^{2}