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After winning the highest place in the B. A examination Sri Venkataraman appeared for 10 the M. He had a superb retentive memory. Basically, it is used to subtract any number from a power of ten. The powers of ten include numbers like 10, , , , etc. So if you want to learn a method by which you can quickly subtract a number from a power of ten, then this technique can come to your aid.
When we go to the market to buy something, we generally give a hundred rupee note to the shopkeeper and calculate the change that we should get after deducting the total amount of groceries. In such a situation, this technique can come to our aid.
Q Subtract In this case, we generally start from the right and subtract 6 from 0. But, we realize that it is not possible to subtract 6 from 0 and so we move to the number in the left and then borrow one and give it to zero and make it ten and so on.
This whole procedure is slightly cumbersome and there is a possibility of making a mistake too. Vedic Mathematics provides a very simple alternative. This will give us the answer. The number to be subtracted is We have to subtract all the digits from nine except for the last digit which will be subtracted from ten.
Thus, The final answer is The respective answers will be 6, 5, 2, 1, 7, 7, 1 and 9. Thus, the final answer is In examples d and e we added zeros in the number below so that we get the accurate answer.
In this chapter we have seen 6 simple yet quick techniques of Vedic Mathematics. I wanted to begin the book with these easy techniques so that we can prepare ourselves for the comprehensive techniques that will follow in the forthcoming chapters. You must have observed that the techniques that we have employed in this chapter work with a totally different approach.
We have found answers to our questions using a completely different approach. In all the chapters of this book you will discover that the method used in solving the problems is far more efficient than the normal systems that we have been using, thus enabling us to produce outstanding results. The traditional system can also be used for numbers of any length. Let us have a look at an example: This is the traditional way of multiplication which is taught to students in schools.
This system of multiplication is perfect and works for any combination of numbers. In Vedic Mathematics too, we have a similar system but it helps us to get the answer much faster. This system is also a universal system and can be used for any combination of numbers of any length. The applications of this system are manifold, but in this chapter we shall confine our study only to its utility in multiplying numbers.
We shall call it the Criss-Cross system of multiplication. We write 6 in the ones place of the answer. We write the 7 in the tens place of the answer. The completed multiplication is: Let us notate the three steps involved in multiplying a 2-digit number by a 2-digit number. We shall have a look at one more example. We write down 7 in the tens place of the answer and carry over 1. But, we have carried over 1. So, the final answer is 7. The answer is We write down 2 and carry over 1.
The total is Now, we add the 1 which we carried over. So, we write 2 and carry over 1. To it, we add the 1 that is carried over and get the final answer as 3. The completed answer is Thus, we see how the Criss-Cross system of multiplication helps us get our answer in just one line!
And, the astonishing fact is that this same system can be expanded to multiplication of numbers of higher digits too. And in every case, we will be able to get the answer in a single line. Let us have a look at the multiplication process involved in multiplying a three-digit number by another three-digit number.
Thus, the final answer is 4. We add the three answers thus obtained to get the final answer 5. The final answer is 6. Thus, it can be seen that the product obtained by multiplying two 3-digit numbers can be obtained in just one line with the help of the Criss- Cross system. We shall quickly have a look at how to multiply two 3-digit numbers where there is a carry over involved. Obviously, the process of carrying over is the same as we use in normal multiplication. We write down 0 and carry over 2.
The total is 30 and we add the 2 carried over to get We write down 2 and carry over 3. The total is 27 and we add the 3 carried over to get the answer as We write down 0 and carry over 3. The total is 11 and we add the 3 carried over. We write down 4 and carry over 1. Three plus 1 carried over is 4. We can expand the same logic and multiply bigger numbers. Let us have a look at how to multiply 4-digit numbers.
A question may arise regarding multiplication of numbers with an unequal number of digits. Let us suppose you want to multiply by Here, we have one number that has three digits and another number that has four digits. Now, for such a multiplication which technique will you use? Will you use the technique used for multiplying numbers of 3- digits or the technique used for multiplying numbers of 4 digits?
To multiply by , write the number as and then multiply it by Use the technique used for multiplying four-digit numbers. Thus, if we want to multiply by 64, we will write 64 as and then multiply it by using the technique of 3-digit multiplication. The knowledge of these characteristics will help us to easily calculate any answer. To understand the characteristics, we shall carefully observe the steps used in multiplying 2-digit, 3-digit and 4-digit numbers.
Most students will be able to guess the trend of the steps used in this system of multiplication by the mere observation of the examples that we have solved. This is because they follow a systematic pattern. The second characteristic of the Criss-Cross system of multiplication is that the number of steps used will always be an odd number. Amongst these odd number of steps, the first and last steps are mirror images of one another, the second and second-to-last steps are mirror images of one another and so on.
In every case there will be a central step, which will be independent that is, without any corresponding mirror image. In our daily practice, we mainly deal with small numbers of 2, 3 and 4-digits. Thus, I have given special emphasis to them. However, to test whether you have understood the logic of this system, I suggest that in the empty space provided next, you notate the steps that we will require if we were to multiply 5-digit numbers. If you got all the steps correct then it is indeed praiseworthy.
Those who made a mistake in other steps need not get disheartened: with enough practice you shall be able to solve sums effortlessly. I would again like to draw your attention to the trend of the steps used in the multiplication procedure. Since we are multiplying two five-digit numbers, the number of steps according to the formulae — 2 multiplied by the number of digits minus 1 — will be 9, and thus we have used steps from a to i.
Further, you will observe that steps a and i are mirror-images, steps b and h are mirror-images, steps c and g are mirror images, steps d and f are mirror images and step e is a central figure. The Criss-Cross system not only helps us to get answers quickly but also helps us to eliminate all the intermediate steps used in the multiplication process. This quality of the system can be of immense aid to students giving various competitive and professional exams.
Let us take a hypothetical situation: We know that in competitive exams, we are given a question with four alternatives out of which we have to select the correct one. Q What is the product of by ? Now, I read the question and start the multiplication process using the Criss-Cross system. First, I multiply the extreme digits 1 by 2 and get the final answer as 2.
My final answer is 4. The moment I get the digits 4 and 2 as the last two digits of my answer, I discontinue the multiplication process and instantly tick option c as the correct answer to the problem. The reason is that the last two digits of the other alternatives are 22, 92 and 44 but I know the last two digits of my answer are 42 and so the correct answer can only be option c.
This was just an example. The idea which I am trying to convey is that while solving any multiplication problem, the moment you get a part of the answer which is unique to any one of the given alternatives, you can instantly mark that alternative as the correct answer. This advantage is not available with the traditional system as it compels you to do the whole multiplication procedure with the intermediary steps.
This comparison proves the fact that the urdhva-tiryak sutra which I call the Criss-Cross system helps us to solve any multiplication problem instantly. There are numerous methods in mathematics which help you to multiply numbers quickly but most of them can be used only for a particular category of numbers, like numbers adding up to 10, numbers between 11 and 20, etc.
However, this system is a universal system and can be used for any combination of numbers. In the next page, I have given the steps for multiplying six-digit and seven-digit numbers. In this manner we can go on and on and on, with eight-digit, nine-digit, ten-digit numbers, etc. The concept that we have studied here can be applied in multiplying algebraic expressions too. For solved examples refer to Appendix B.
Many of these techniques have their roots in multiplication as squaring is simply a process of multiplication. The reason for its popularity is that it can be used for any type of numbers. Ex: Find the square of 23 a First, we multiply 3 by 3 and get the answer as 9. The answer is 5.
The final answer is Similarly, numbers of higher orders can be squared. Refer to the chapter on Criss-Cross system for further reference. Let us discuss them one by one. This method is generally to square numbers which are near multiples of It is a part of the regular school curriculum. The formula is This formula is very much like the first one. The only difference is that the middle term carries a negative sign in this formula.
Q Find the square of We will express the number as — 5 Q Find the square of We will express the number as - 2 Thus, we see that the two formulae can help us find the squares of any number above and below a round figure respectively. There is another formulae which is used to find the square of numbers, but it is not so popular. I discuss it below.
However, if one finds multiplying 74 by 70 difficult, we can simplify it still further. First, multiply 70 by 70 and then multiply 4 by 70 and add both for the answer. Because of this, the value of a - b became 70 and the multiplication procedure became easy as the number 70 ends with a zero.
Calculating the cube root of a number by the traditional method is a slightly cumbersome procedure, but the technique used by Vedic Mathematicians is so fast that one can get the answer in two to three seconds! The technique for solving cube roots is simply so amazing that the student will be able to correctly predict the cube root of a number just by looking at it and without the need for any intermediate steps. You might find it difficult to believe, but at the end of this study, you will be calculating cube roots of complicated numbers like , and in seconds.
Even primary school students who have learnt these techniques from me are able to calculate cube-roots in a matter of seconds. Before we delve deeper in this study, let us clear our concepts relating to cube roots. Let us take the number 3. When we multiply 3 by itself we are said to have squared the number 3.
When we multiply 4 by itself we are said to have squared 4 and thus 16 is the square of 4. Similarly the square of 5 is 25 represented as 52 The square of 6 is 36 represented as 62 In squaring, we multiply a number by itself, but in cubing we multiply a number by itself and then multiply the answer by the original number once again.
Now, since you have understood what cubing is it will be easy to understand what cube rooting is. Cube-rooting is the procedure of determining the number which has been twice multiplied by itself to obtain the cube. Calculating the cube-root is the reciprocal procedure of calculating a cube. Thus, if 8 is the cube of 2, then 2 is the cube-root of 8.
If 27 is the cube of 3 then 3 is the cube root of 27 And so on. In this chapter, we will learn how to calculate cube-roots. Thus, if you are given the number 8 you will have to arrive at the number 2. If you are given the number 27 you will have to arrive at 3. However, these are very basic examples.
We shall be cracking higher order numbers like , , etc. At this point, I would like to make a note that the technique provided in this chapter can be used to find the cube-roots of perfect cubes only. It cannot be used to find the cube root of imperfect cubes. This list will be used for calculating the cube roots of higher order numbers. With the knowledge of these numbers, we shall be able to solve the cube-roots instantly.
Hence, I urge the reader to memorize the list given below before proceeding ahead with the chapter. The cube of 1 is 1, the cube of 2 is 8, the cube of 3 is 27 and so on…. Once you have memorized the list I would like to draw your attention to the underlined numbers in the key. You will notice that I have underlined certain numbers in the key. These underlined numbers have a unique relationship amongst themselves.
In the first row, the underlined numbers are 1 and 1. It establishes a certain relationship that if the last digit of the cube is 1 then the last digit of the cube root is also 1. In the second row, the underlined numbers are 2 and 8. It establishes a relationship that if the last digit of the cube is 8 then the last digit of the cube root is 2. Thus, in any given cube if the last digit of the number is 8 the last digit of its cube-root will always be 2.
In the third row, the underlined numbers are 3 and 7 out of 27 we are interested in the last digit only and hence we have underlined only 7. We can thus conclude that if the last digit of a cube is 7 the last digit of the cube root is 3. And like this if we observe the last row where the last digit of 10 is 0 and the last digit of is also 0. Thus, when a cube ends in 0 the cube-root also ends in 0. There is one more thing to be kept in mind before solving cube-roots: Whenever a cube is given to you to calculate its cube-root, you must put a slash before the last three digits.
If the cube given to you is you will represent it as If the cube given to you is , you will represent it as 39 Immaterial of the number of digits in the cube, you will always put a slash before the last three digits.
First, we shall solve the right hand part of the answer and then we shall solve the left hand part of the answer. If you wish you can solve the left hand Part-Before the right hand part. There is no restriction on either method but generally people prefer to solve the right hand part first. As illustrative examples, we shall take four different cubes.
We have thus got the right hand part of our answer. In this case, the number lying to the left of the slash is Now, we need to find two perfect cubes between which the number lies in the number line. From the key, we find that lies between the perfect cubes the cube of 6 and the cube of 7. Our final answer is Thus, 66 is the cube root of You will observe that as we proceeded with the examples, we took much less time to solve the cube-roots.
After some practice you will be able to solve the cube-roots by a mere observation of the cube and without the necessity of doing any intermediary steps.
It must be noted that immaterial of the number of digits in the cube, the procedure for solving them is the same. Q Find the cube root of We will put 3 as the right hand part of the answer. We may thus conclude that there exists only one common procedure for solving all types of perfect cube-roots. In my seminars, the participants often ask what is the procedure of solving cube roots of numbers having more than 6 digits. All the examples that we have solved before had 6 or fewer digits.
Well the answer to this question is that the procedure for solving the problem is the same. The only difference in this case is that you will be expanding the number line. Let us take an example. We know that the cube of 9 is and the cube of 10 is Now let us go a step ahead and include the higher numbers.
We know that the cube of 11 is and the cube of 12 is Out of 11 and 12 the smaller number is 11 which we will put beside the 2 already obtained. Hence, the final answer is The two examples mentioned above were just for explanation purposes. Under normal circumstances, you will be asked to deal with cubes of 6 or less than 6 digits in your exams. Hence, knowledge of the key which contains cubes of numbers from 1 to 10 is more than sufficient.
However, since we have dealt with advanced level problems also, you are well equipped to deal with any kind of situation. In the traditional method of calculating cube-roots we use prime numbers as divisors. Prime numbers include numbers like 2, 3, 5, 7, 11, 13 and so on. Let us say you want to find the cube root of Then, the process of calculating the cube root of 64 is as explained below. First, we divide the number 64 by 2 and get the answer It can be represented as: Hence, 4 is the cube root of Similarly, to find the cube root of by the traditional method, we can use the following procedure.
Hence, 15 is the cube-root of After studying the above two examples, the reader will agree with me that the traditional method is cumbersome and time-consuming compared to the method used by Vedic mathematicians. However, you will be shocked to see the difference between the two methods when we try to calculate the cube root of some complicated number.
This characteristic of this system helps students in instantly cracking such problems in competitive exams.
With this comparison we terminate this study. Students are urged to solve the practice exercise before proceeding to the next chapter. Most school and college exams ask the square roots of perfect squares.
Therefore, this chapter is very useful to students giving such exams. Students of higher classes and other researchers will find the thirteenth chapter useful as they will be able to study the Vedic Mathematics approach to calculate square roots of any given number — perfect as well as imperfect. The need to find perfect square roots arises in solving linear equations, quadratic equations and factorizing equations. Solving square roots is also useful in geometry while dealing with the area, perimeter, etc.
The concepts of this chapter will also be useful in dealing with the applications of the Theorem of Pythagoras. The technique of finding square roots of perfect squares is similar to the technique of finding the cube root of perfect cubes. However, the former has an additional step and hence it is discussed after having dealt with cube roots.
Squaring of a number can be defined as multiplying a number by itself. The symbol of square is represented by putting a small 2 above the number.
Similarly, 25 is the square of 5, and 5 is the square-root of The squares are given below. Memorize them before proceeding ahead. In the chapter dealing with perfect cube roots we observed that if the last digit of the cube is 1 the last digit of the cube root is also 1.
If the last digit of the cube is 2 then the last digit of the cube-root is 8 and so on. Thus, for every number there was a unique corresponding number. However in square roots we have more than one possibility for every number.
Look at the first row. Here, we have 1 in the number column and 1 in the square column. Similarly, in the ninth row we have 1 in the number column and 1 of 81 in the square column. Do not worry if you do not follow this immediately. You may glance at the table below as you read these explanations, then all will be clear. Similarly, whenever we come across a square whose last digit is 6, we can conclude that the last digit of the square root will be 4 or 6 and so on….
Now, I want you to look at the column in the left. Note that the numbers 2, 3, 7 and 8 are absent in the column. That means there is no perfect square which ends with the numbers 2, 3, 7 or 8.
However, in many cases we will have two possibilities out of which one is correct. Further, we do not know how to find the remaining digits of the square root. So we will solve a few examples and observe the technique used to find the complete square root. Before proceeding ahead with the examples, I have given below a list of the squares of numbers which are multiples of 10 up to This table will help us to easily determine the square roots. Therefore the square root ends with 2 or 8.
We find that the number lies between which is the square of 80 and which is the square of The number lies between and Therefore, the square root of lies between the numbers 80 and From the second step we know that the square root lies between 80 and Of all the numbers between 80 and 90 81, 82, 83, 84, 85, 86, 87, 88, 89 the only numbers ending with 2 or 8 are 82 or Thus, out of 82 or 88, one is the correct answer. If the number is closer to the smaller number then take the smaller number 82 as the answer.
However, if it is closer to the bigger number , then take 88 as the answer. In this case, we observe that is closer to the bigger number and hence we take 88 as the answer. The square root of is Thus, our answer lies between 90 and So, we can eliminate the numbers that do not end with a 1 or 9.
Lastly, we know that the number is closer to the bigger number 10, and so we take the bigger number 99 as the answer. So the square root is between 70 and So, the root lies between 40 and
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