Trick to identify exact squares

While vedic math tricks would make squaring numbers very easy, these few tricks help identify perfect squares merely by observing the number.:


Structure of the square root:
First trick is to understand what would be the structure of the square root just by observation.

• The number of digits before decimal point would reduce to half when we take the square root of the number.(i.e. the square root of 1446.78 would have 2 digits before decimal number). This math trick would be applicable directly when the numbers of digits before decimal point are even.
• In case of odd number of digits before decimal point, the square root would still have half of the numbers but that half needs to be rounded up to the next integer. (i.e. square root of 144.678 would have two digits before the decimal point.)
• The number of digits after the decimal point would reduce to half when we take the square root of the number.(i.e. the square root of 1446.78 would have 1 digits after decimal number). This math trick would be applicable directly when the numbers of digits before decimal point are even.
• For numbers with odd digits after decimal point, the square root would be an irrational number, because any square of a whole rational number would have even digits only after the decimal point.

Nature of the square root:Next trick is to know the type of answer to expect while trying to derive square root of a number, the rule is:

• For the numbers not exact squares of another whole number, their square root is always irrational. (i.e. the square root would have endless digits after decimal point and would not have any repeating patters in those digits.)
• Also, a number is odd number of zero's at the end, would never have a rational square root
• For the rest of the numbers the square root would be rational.

The last digit of the answer:This is a trick to identity an exact square and also to estimate the last digit of the square root.

• Number with last digits as 2, 3, 7 or 8 are not exact squares and inturn would result in an irrational number when their square root is calculated
• Also, there is a direct relation between last digit of the number and last digit of its square root, as follows:
• number 1, square root 1 or 9
• number 4, square root 2 or 8
• number 5, square root 5
• number 6, square root 4 or 6
• number 9, square root 3 or 7
• When the second last digit (from right) is even and the last digit is 6, the number is not a perfect square (i.e. 346 is not a perfect square)
• When the second last digit (from right) is odd, the last digit has to be 6 for the number to be a perfect square. (i.e. numbers ending with 34, 59, 11 are not perfect squares)
• When a number is even, and its last two digits taken together are not divisible by 4, that number is not a perfect square. (i.e. numbers ending with 42, 86, etc. are not perfect squares)

Odd one out:This trick will tell us whether the square root would be even or odd.

1. Square root is odd when a perfect square is odd
2. Square root is even when a perfect square is even

These tricks would be particularly helpful when attempting multiple choice questions or math puzzles relating to squares or square roots.

Finding Squares

You know the squares of 30, 40, 50, 60 etc.

but if you are required to calculate square of 31 or say 61 then you will scribble on paper and try to answer the question.

Can it be done mentally?

Some of you will say may be and some of you will say may not be.

But if I give you a formula then all of you will say, yes! it can be.

What is that formula….. The formula is simple and the application is simpler.

Say you know 60sq = 3600

Then 61sq will be given by the following

61sq = 60sq + (60 + 61) = 3600 + 121 = 3721

or Say you know 25sq = 625 then 26sq = 625 + (25 + 26) = 676

Like above, you can find out square of a number that is one less than the number whose square is known.

Amazing way to calculate squares within seconds

"Ekadhikena Purveṇa" as we discussed for division is the Sanskrit term for "One more than the previous one" and very important Vedic Mathematics formula. We already know from the previous post that this formula can be used for both multiplication and division, here let us have a look at how it helps for multiplication:
Application:
This formula can be applied to multiplication of numbers that satisfy both the following conditions

1. the same first digit
2. the sum of their last unit digits is 10.

Even though it might so appear that this formula is useful in limited cased but an interesting category of numbers that satisfy both above requirements are all the numbers ending with 5. So when we need to derive squares for any number ending with 5, this formula is applicable straightaway. In this post, I am going to talk about limited application of this formula for deriving squares of numbers with last digit as 5.

Method:

Let us break the method into two parts:

1. Deriving last two digits of answer
2. Deriving all previous digits of answer

1. Deriving last two digits of answer

No application of mind, last two digits are 25, always.

2. Deriving all previous digits of answer

All the digits in the number before the last five would constitute base for us and let us call them B1.

Another base is B1 +1 (Because the formula is about one more than the previous one…)

Digits previous of 25 in the final answer would be simply [B1 x (B1+1)]

Example:

Let us understand this with an example of squaring 25

Last two digit of the answer are 25

For other digits, we need B1 and it goes as follows:

B1 = 2 (All the digits other than the last 5 in the number we want to derive square for)

B1 + 1 = 2 + 1 = 3

Therefore other digits of the answer preceding 25 are = 2 x 3 = 6

Hence the final answer would be 625.

So now squaring is not a headache anymore having understood vedic mathematics, right?

Logic:

Learning just the shortcut method isn’t enough, so let us understand the logic how this operates. For explaining the logic we need to use the basic concepts of factorization:

We all know that (a+b)² = a² + 2ab + b² 

Now how does this square of numbers ending with 5 work in this case?

Let us continue with the above example:

25²

(20 + 5)²

20² + (2x20x5) + 5² (Factorization using the above rule)

20² + (20x10) + 5² (Simplified with multiplication of 5x2 for middle term)

20 (20 + 10) + 5² (taking 20 common from first two terms)

20 (30) + 5²

2 x 10 x 3 x 10 + 5²

(2x3) x (10x10) + 5²

(2x3) x (100) + 5²

Now this expression looks like this [B1 x (B1+1)] x 100 + 5²

Because in this expression two things are constant i.e. whatever is the B’s multiplication that shall at the end be multiplied by 100 and 25 shall be added so it is safe to make a rule that last two digits of the answer are going to be 25 and right preceding these digits we need to insert the multiplication of B1 & B1+1

Now you must feel awakened with the logical explanation of Vedic mathematics.