Cool Math Tricks

After you are through with these tricks you will be hailed as a math genius.

These cool math tricks would help parents help their kids getting up to speed in learning math. Also, a nice and funny way to impart education in your children and pamper their curiosity for numbers.

These cool math tricks, derivation of various vedic math sutras, make calculations possible at your fingertips. Please feel free to leave your comments in case you have any suggestions for additions to this list of cool math tricks! These cool math tricks are particularly important to those appearing and preparing for competitive exams like GMAT, GRE, etc.

Trick to multiply with 11
Multiply any number with 11 easy and fast 

This cool math trick is about multiplication. Not just any number with any other. This math trick has specific application and its about multiplication of any number with 11. With this trick, multiplication of any number with 11 can be calculated in two seconds. Let us see how this trick works with an example of two digit number to be multiplied with 11:

Example:

24 x 11 (Since this math trick is about multiplication with 11, the number we want to multiply 11 with, will be referred to as "Number")

2 _ 4 (The number is split in half from the middle with space in between)

2 6 4 (In the space left in between, total of the two digits on either sides is to be included)

24 x 11 = 264


Cool Math Trick to multiply with 9
use your fingers to multiply any number with 9

This cool math trick is a simple exercise and kids will love this cool math trick. Let us assume, you want to multiply 2 by 9 and don't want to use any of the mathematics concepts. This math trick would let you do it, like this:

1. Raise your both hands up and place them on a platform (anything in front of you, even the desk on which this computer is placed would do). Also, please keep your fingers spread, because that only will make the difference. 
2. As you want to multiply with 2, you just need to fold down the second finger from the left. And just watch your resulting fingers. They would give you the answer.
3. There would be only one finger to the left of folded finger, which signifies that left hand side digit of the answer would be 1. 
4. There are eight finger to the right of folded one, which means the right hand side digit of the answer is 8. (3 of left hand and 5 of right hand) 

Please try this on for any number and this math trick would give you exact right answer for multiplication with 9.

Isn't it cool to know such math tricks! Kids, enjoy playing with these cool math tricks!

Magic with mathematics
let us explore the magic of 1089

This cool math trick would turn you into a magician (Shh! only till your friends also know this :-))

Here is the math trick you need to play with your friends. Why dont you play along as you read. Take a pencil and paper and let us begin with this cool math trick.

I can guess the final answer of all the following steps for any number your guess. And the answer would be 1089!

Isn't it a cool math trick? Enjoy fooling your friends and also think about the reason why any three digit number if passed through following fives steps would give the same result?

The hint to understand logic behind this cool math trick is - "when you reverse a three digit number, tenths place digit would be unmoved and the hundredth and unit would change places"

1. Please write down any three digits number with decreasing digits
   (i.e. 543 or 875, you can also use the number which have zeroes in their tenth or unit place)
2. Reverse the number you wrote in #1
   (If you presumed number with zeros either in tenth or unit or both places, the number you would get in this step would not necessarily be a three digit number)
3. Subtract the number obtained in #2 from the number you wrote in #1
   (Please note, because number in #1 was with digits decreasing, the number got in #2 would be lower)
4. Reverse the number obtained in #3
   (The digits of the subtraction received in #3 to be written in reverse order. Please note, in both the reversals, tenths place digit would be unmoved and the hundredth and unit would change places)
5. Add the numbers found in #3 and #4

Mental Multiplication

Here we are talking about the sutra "Nikhilam" which would help us solve the mathematical multiplication in our mind without the use of calculator and even without having to write anything down! Let us understand the method with the help of examples.



Example 1:

605 x 504 = ?

We first need to select a base number (preferably a multiplier of 100 in case of three digit numbers to make it easier). In this case the base would be 500.

Now, define the distance of multipliers from the base as given below

605 -500 = +105 (the sign is very important)

504 -500 = +004

Now lets reproduce these numbers as:

605 + 105

504 + 004

From this, we derive two numbers and their names would be as follows:

LHS (is the result of multiplying the deficiencies) = +105 x +104 = +420

RHS (by adding diagonally on any one side) = 105 + 504 = 609 (remember, you can select any diagonal side, the answer will remain the same)

Now because our base is not a 10 series number (i.e. 10, 100, 1000, 10000, etc.) we need to establish the relation of the base with next highest 10 series number and the relation is 1000 / 2 = 500

Therefore, we have to express the LHS and RHS as follows:

(609 /2) | (+420)

304.5 + 420

304.5 | 420

Now see that the L.H.S has a decimal point ,so carry out that half to the R.H.S side from L.H.S.

Now R.H.S side becomes 420 + 500 = 920

304 | 920

Therefore ,605 x 504 = 304920

Example 2

485 x 475 = ?

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above both numbers are less than the base.

How much deficient are are the above numbers are from base ?

485 -500 = -015

475 -500 = -025

485 - 015

475- 025

--------------
460 | ( +375) -->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

230 | 375 ---- (Half of 460 is 230)

230 | 375

Therefore 485 x 475 = 230375

Example 3

614 x 495 =?.

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above numbers .One is greater than the base and one is less than the base.

How much deficient are are the above numbers are from base ?

614 - 500 = +114

495 - 500 = -005

614 + 114

495 - 005

------------
609 | (-570)--->-->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

(609 /2) | (-570)

304.5 | (-570) ----- > the minus sign is removed by 1000 -570 =430 and remove one from the L.H.S side

303.5 | 430

303| (430+500) ----> the 0.5 when carried to R.H.S, becomes 500 and add it to R.H.S

303| 930

Therefore ,614 x 495= 303930

These are general rules. There would not be issues when multiplications are above or higher, but when one is greater and one is lesser, follow the rules in the last example. All in all, the rules are:

The (0.5) on L.H.S is carried as 500 and added to the R.H.S every time you fall into a decimal (1/2).
The Minus signs on R.H.S is removed by adding 1000 to it.

It would seem very easy once you get in this mode of calculations.

Multiplication at your fingertips!

Here I am going to talk about "Ekadhikena Purveṇa" i.e "One more than the previous one" and very important Vedic Mathematics formula and how to use it 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.

We have already discussed about applying this formula for deriving squares for any number ending with 5, in previous post (Amazing way to derive squares within seconds). So here, I am going to talk about application of this formula for multiplication of numbers satisfying the conditions stated above.

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
These shall be multiplication of last digit of the numbers we are looking to multiply.

2. Deriving all previous digits of answer

All the digits in the multipliers other than the last 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…)

So previous digits in the final answer would be simply [B1 x (B1+1)]

Example:

Let us understand this with an example: 24 x 26

Last two digit of the answer:

4 x 6 = 24 (If this answer is in single digit, then we need to put 0 before it)

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 624.

Logic:

So what’s the logical explanation of this wonderful shortcut:

We all know that (a-b) x (a+b) = a²-b²

Now how does this help with multiplication of 24 & 26? It is like this:

24 x 26

(25-1) x (25+1)

25² – 1²

625 – 1 (25² derived from the same formula applied for squaring discussed in the previous post)

624 (Well this matches!)

Vedic mathematics rocks!

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.

Multiplication Made Easy Using The Line System

Now that we have seen the simpler way of multiplication using vedic mathematics for two digit numbers at Basics of Vedic Mathematics, I will show you an alternate method which even doesn't require multiplication of two single digit numbers to derive product of any large numbers.

Keep connected to www.gyancircle.com for more updates.

This is the fun of vedic mathematics where you can multiply without multiplying!

Basics of Vedic Mathematics

After introduction to Vedic Mathematics, let us start with basics thereof.
It would be good to take multiplication, and for that any 2 digit number with any other 2 digit number.

Basic Definitions:

Vedic mathematics is based on the concept of placing the numbers either at the unit place or tenths or hundredth and so on. So for ease of understanding, let us use this legend:

Unit Place: UP

Tenth Place: XP

Hundredth Place: HP

Thousand Place: TP

Ten thousandth Placce: TXP

Hundred thousandth Place: THP (and so on..)

Let us take an example of 23 x 45:

So now we need to place this numbers under each other in such a way that UP of both multipliers are in one column and XP numbers in one column:

XP UP

2 3

4 5

1. Now we have to start with UP and multiply both the numbers there. (i.e. 3 x 5) and the answer is 15. Out of this answer 5 would be placed on the UP of the product of two multipliers and 1 would be a carry over. So we now know that our final product has a UP of 5.

2. Second step is then to go for cross multiplication which means:

"XP of first multiplier" x "UP of Second multiplier" = 2 x 5 = 10

"XP of Second multiplier" x "UP of First multiplier" = 4 x 3 = 12

3. Now the sum of above two products added with carry over if any, would give us the XP of our final answer which is to be derived as follows:

Sum of Cross Multiplication (from Step 2) = 10 + 12 = 22

Carry over from the UP (from Step 1) = 1

XP for the final product = 23

As we did in step 1, number 3 shall occupy the XP of final product and 2 shall be a carry over. so the answer constructed by us so far look like this _ _ 3 5. Now we shall go into final steps to identify TP and TXP of the final product.

4. Multiplication of XP of both multipliers i.e. 2 x 4 = 8.

We need to add the carry over of two from step 3 into this which would give us 8 + 2 = 10. So 0 is the TXP and 1 is TP of the final product.

Thus, the final product looks like this 1035.

Benefits of this method

Now let us revisit and see what we have actually done to achieve this product of two digit numbers:

1. Multiplication of single digit number

2. Sum of two digit numbers.

Thus, the requirement of multiplying two digit numbers has become very very easy using this Vedic Maths technique and with a little bit of practice on this lines would make you very quick in deriving products.

Concept

Also, to understand the concept better, let us take this pictorial demonstration of steps. one this is understood, you shall be able to apply the concept to even larger digit numbers as well.


So Vedic Maths Rocks!

I hope this would help you master the trick for multiplication of two digit numbers and shortly I shall come up with multiplication of larger digit numbers using wonders of Vedic Mathematics.

As I mentioned previously, in this attempt to simplify the study and make it more interesting, I am supported by Gyancircle.com. Do visit them.