This system of multiplication is adopted from Vedic Mathematics' URDHVA-TIRYAK SUTRA, which means vertically and cross-wise.

To start with, we will look at a simple example just to get a grasp on the steps involved in the method. Later we will apply it to a slightly more advanced problem to show how to handle carrying numbers from one digit to the next. For now, we will multiply 111 by 111.

111

x 111

First, you will take the right-hand digits and multiply them together. This will give you the one's digit of the answer, as shown below with the digits used encased in brackets.

11[1]

x11[1]

____

* - -- 1

Next, multiply the one's digit of the top number by the ten's digit of the bottom number, and the one's digit of the bottom number by the ten's digit of the top number. Once you have those values, add them together, and you will have the ten's digit of the answer. The digits you multiply together are enclosed in the same type of bracket. This gives (1*1)+(1*1), so the ten's digit is equal to two.

1{1}[1]

x 1[1]{1}

______

* 321

For the next step, all digits of the number will be involved in order to find the middle of the answer. Multiply the one's digit of one to the hundreds digit of the other, and then multiply the ten's digit of both together, then finally add them all together. This will give you (1*1)+(1*1)+(1*1) for a value of 3. As above, the digits paired together are enclosed in the same type of bracket.

{1}[1]

x [1]{1}

_______

* 321

The fourth step is similar to the second step, just moved one place to the left. You multiply the ten's digit of one number by the hundred's digit of the other number. Again you will get (1*1) + (1*1), showing the thousand's digit of the answer is equal to 2.

{1}[1]1

x {1}[1]1

______

* 2321

For the final step, simply multiply the left-hand number of both numbers together to get a value of 1 for the ten-thousand's place.

{1}11

x {1}11

____

12321

To give a real-world example, consider that you want to buy 15 of some product for $1.25 each. You can consider 15 to be a three digit number, where the third digit is equal to 0.

$1.25

x 015

_____

* - -- -

You will notice that immediately you will have to deal with carrying over value from one digit to the other. This works very similar to regular multiplication methods. You simply take any value in the tens digit of that step as an addition to the next step. When you multiply 5 by 5 in this example to get 25, you would place the 5 as the one's digit of the answer, and add the 2 to the next step to find the ten's digit. This makes the second step equal to (5*2) + (1*5) +2, for a total of 17. Use the 7 as the value for the ten's digit, and carry the one over to the next step. Here you would end up with (5*1) + (1*2) + (0*5) + 1, coming to a value of 8. This time there is no number to carry over, so proceed through the rest of the problem as normal. The thousands place is 1 via (0*2)+(1*1), and the ten-thousands place is equal to 0 via 0*1. This comes to 01875, then drop the 0 from the end.

As in regular multiplication, you count the total number of places behind the decimal point, and add the same number to the answer. This means the items would cost $18.75 to purchase.

Justin McGenity is a freelance writer, science geek, and self-proclaimed philosopher. His hobbies include studying, video games, and socializing. You can find him writing for such sites as Degree Jungle a resource for university students.

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Finding Last Digit of Any Number Raised to Any Power

"Hello sir , I like your shortcuts very much. But please tell me how to find unit’s digit in numbers like 7 raise to power 205, 19^239. Sir I am pissed off solving these type of problems"

- Except of the comment on an earlier post - Shortcut Method for Multiplication

Cyclicity Explained 

To understand cyclicity let us take a simple example.

Take any two numbers say 43 and 97.

If they are multiplied, the answer is 4171. The last digit of the product is same as the last digit of 3 x 7.

Hence, it is 1.

This concept could be extended to a host of situations. An interesting pattern emerges when we look at the exponents of the numbers. We would find conclusions as given below.

The last digits of the exponents of all numbers have cyclicity i.e. every Nth power of the base shall have the same last digit, if N is the cyclicity of the number. All numbers ending with 2, 3, 7, 8 have a cyclicity of 4.

For instance,

2^1 ends with 2

2^2 ends with 4

2^3 ends with 8

2^4 ends with 6

2^5 end with 2 again.

The same set of the last digits shall be repeated for the subsequent powers. So, if we want to find the last digit of (say) 2^45, divide 45 by 4.

The remainder is 1

So the last digit would be the same as last digit of 2^1, which is 2

Let us take a CAT level example

The digit in the unit place of the number represented by (7^95 * 3^58) is

A. 7
B. 0
C. 6
D. 4

Answer: A (7)

Solution

Cycle of 7 is

7 ¹=7

7 ²=49

7 ³= 343

7 ⁴= 2401

If we divide 95 by 4, the remainder will be 3.

So the last digit of (7)⁹⁵ is equals to the last digit of (7)³ i.e. 3.

Cycle of 3 is

3¹ =3

3² =9

3³= 27

3⁴= 81

3⁵= 243

If we divide 58 by 4, the remainder will be 2. Hence the last digit will be 9.

Therefore, unit's digit of (7^95 * 3^58) is unit's digit of product of digit at unit's place of 7^95 and 3^58 = 3 * 7 = 21. Hence 1 is the answer.

Working out similarly for all other digits we get

CYCLICITY TABLE

Using the above table try answering the questions raised in the comment by a QuickerMaths follower, re-posted above.

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You are in a room with 2 doors. Behind 1 door is a coffer overflowing with jewels and gold, along with an exit. Behind the other door is an huge, hungry lion that will pounce on anyone opening the door. You have no clue which door leads to the treasure and exit, and which door leads to the lion.

In the room there are 2 more individuals. The first is a Sachcha, who always tells the truth, and the second is a Jhoota, who always lies. Both of these individuals know what is behind each door. You do not know which individual is the Sachcha, or which one is the Jhoota. You may ask one of the individuals exactly 1 question. What should you ask in order to be certain that you will open the door with the coffer behind it, instead of the hungry lion?

Post your solution as a comment below -

If you like such logic puzzles, you can explore more by visiting -

Logic Puzzle of Knights and Knaves

Logic Puzzle of Twin Brothers

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We can easily calculate the square of any number ending in 9 using the method described in this post. Let us understand this method with the help of an example –

Finding the square of 39

Firstly add 1 to the number. The number now ends in zero and is easy to square.
40^2 = (4*4*10*10) = 1600. This is our subtotal.

In the next step, add 40 plus 39 (the number we squared plus the number we want to square)

40 + 39 = 79

Subtract 79 from 1600 to get an answer of 1521. To easily do such subtractions, subtract 80 (1 more than 79) from 1600 to get 1520 and then add 1 to get the answer as 1521.

1600 – 79 = 1521 is the square of 39. Answer.

Another Example of finding the square of bigger number ending in 9.

Finding the square of 159

159^2 =
159 + 1 = 160
160^2 = 25600
160 + 159 = 319
25600 – 319 = 25600 – 340 + 1
25600 – 319 = 25281

Another Example,

449^2
450^2 = 202500 (use the shortcut Squaring number ending in 5 to calculate 45^2 and then put double zeroes in front of the answer)
450+449 = 899
202500 – 899 = 202500 – 1000 + 101 = 201601

To make sure that you have understood the above calculation shortcut, try finding the square of few numbers ending in 9 by yourself – 39, 119, 349.

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By this method we can find out Sines and Cosines of different angles. It just requires your hand. Let’s understand this trick step by step -

Step 1

First mark the angles of 0, 30, 45, 60, and 90 on little, ring, middle and pointer finger and thumb of your left hand.

Step 2

On the palm of your left hand write the equation (x)^1/2 /2 or  square root of x/2

Step 3

On the left will be cosine and on the right will be sine. You will understand why you need to do this in the next step with an example

Step 4

Suppose, you want to find Cos30

Fold the finger representing 30. i.e. ring finger of your left hand palm.

Count the numbers of fingers on the left of the ring finger, as per step 3. So since there are 3 fingers, x=3; put the value in the equation given in step 2

Hence, Cos30 = square root of 3/2

and Sin30 will be (sq. root of 1 /2) = 1/2

I appreciate the efforts of Debasis in sharing this trick with all of us. In case you want to share some quick calculation trick or technique, mail it to me at vineetpatawari [at] gmail [dot] com

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Vedic Mathematics – This book throws light on the true knowledge of Vedic mathematics which relates to the truth of numbers and magnitude, applicable to any background- either from arts or science. After eight years of intensive effort of Jagadguru Sri Bharati Krsna Tirtha this volume, more a 'magic', is the result of intuitive visualization of fundamental mathematical truths. The ancient Indian methods are analyzed and shown to be capable of solving various problems of mathematics. This book contains forty chapters which deal with all the subjects in mathematics- Multiplication, Division, Factorization, Equations, Calculus, Analytical Conics, etc. - Purchase Online

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In any two dimensional figure, if the corresponding sides are in the ratio a:b, then their areas are in the ratio a²:b²

Two dimensional figures can be any polygon like square, rectangle, rhombus, trapezium, hexagon, etc. It can also be a triangle or a circle. The sides, referred in the statement above, can be length, breadth or even diagonal in case of a polygon. In case of a circle the sides will be represented by radius or diameter or circumference. In triangle it can be sides or height of a triangle.

To understand the above concept let us take few examples:

Problem: The sides of a hexagon are enlarged by three times. Find the ratio of the areas of the new and old hexagon.

To solve this problem students normally assume the sides of smaller hexagon as x. Hence the corresponding sides of the enlarged hexagon become 3x. Then they calculate areas of respective hexagon using formula, Area = (3√3*x²)/2. Then they will find the ratio of areas of both the hexagons.

Using the shortcut above, we know ratio of the corresponding sides of the two hexagons is a:b = 1:3

Therefore, ratio of their areas is given by a²:b² = 1²:3² = 1:9

Problem: The ratio of the diagonal of two squares is 2:1. Find the ratio of their areas.

Using the above shortcut, the ratio of their area will be 2²:1² = 4:1

Problem: The ratio of the radius (or diameter or circumference) of two circles is 5:7. Find the ratio of their areas.

Using the above shortcut, the ratio of their area will be 5²:7² = 25:49

The same logic can be extended to any 3-dimensional figure like cube or sphere or cone or any other.

In any two 3-dimensional figures, if the corresponding sides or other measuring lengths are in the ratio a:b, then their surface area are in the ratio a²:b² andvolumes are in the ratio a³:b³

Problem: The sides of two cubes are in the ratio 2:1, find the ratio of their surface area and volumes.

The ratio of their surface area is 2²:1² = 4:1 and ratio of volumes would be 2³:1³ = 8:1

Problem: The radiuses of two spheres are in the ratio 3:4, find the ratio of their surface area and volumes.

The ratio of their surface area is 3²:4² = 9:16 and ratio of volumes would be 3³:4³ = 27:64

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Ramlal is a night watchman in a large company. On one fine morning when Ramlal was about to leave for home, his boss informs him, “I'll go for a business trip to Colombo. Tomorrow I will depart from Chennai airport”.

Ramlal, however advises him to take a ship.

“Why should I?” inquired the boss.

“Yesterday night I dreamt that the plane to Colombo crashes, just before it will land”, is the response from Ramlal.

The president smiles first, but since he is pretty superstitious he decides to take the ship. When he arrives in Colombo, he is told that the plane which he should have taken had crashed. When the president returns from the trip, he gives a big reward to Ramlal and immediately fires him. Why?

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The links above are affiliate links to Flipkart. I would request you to read the reviews and take opinion of friends and teachers before ordering any book. I shall be glad to add books suggested by you, to the above list. Please feel free to add your suggestions by posting a comment below.

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Carrying over is a concept taught to us at a very early stage of our life. However, it has never been an easy thing to do. In adding two or more numbers, most of us face problem while “carrying over”. Larger the digits, involved in the numbers to be added, more likely it is to involve carrying.   More the carrying over involved, more likely are we to make mistakes.

Friends, remember the most basic and effective rule of making arithmetic fast and quick is to break difficult calculations into simpler, easily manageable small calculations.

Any digit when added to 9 (except 0) makes carrying over mandatory.  On the contrary, anything added to 0 can’t produce a two digit number.  Even if 9 is added to 0, no carrying is needed.

So, to make things simpler let us create some zeroes.

Example:

Add 38 + 86

First make 38 to 40 by adding 2.  Now obviously adding 86 to 40 is definitely easier than adding 86 to 38.

86 + 40 = 80 + 40 + 6 = 126

I am sure you must be concerned about the 2 we added out of nowhere.  Well you must be, but if you can balance out this extra 2 by subtracting 2 from the answer (126), the final answer will be the same.

You can create 0 towards the end of both the numbers to be added.  Try to understand this with an example,

187 + 139

Add 140 (=139+1) to 190 (=187+3)

140 + 190 = 330

Now deduct back ‘1’ and ‘3’ added to the respective numbers. Hence to balance out subtract 1 and 3 from 330 = 330 – 1 – 3 = 326 is the final answer.

The effect of the above trick can be remarkable. Like any other quick calculation tricks, this also requires a lot of practice to master it.

Check out yourself by adding the following numbers -

37 +54 =?

79 + 23 =?

While adding decimals, this can be a very powerful trick. Try these questions -

12.97 + 1.34 = ?

14.95 + 11.60 = ?

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