6. Trick - Division by numbers ending in 9 (Part I)

The sutra (formula) that empowers me to divide by numbers ending in 9 is again "Ekadhikena Purvena" meaning by one more than the previous one

The division can be done in two ways - by division and by multiplication .

I will take up by multiplication in this article. I will illustrate by taking two examples, 1/19 and 1/7 = 7/49

The numerator in 1/19 is 1. So, we start with 1.

Digit previous to 9 is 1 and one greater than 1 is 2. We will deal with multiplications by 2 only.

Multiply 1 by 2 to get 2

Now, we have 21.

Multiply 2 by 2 to get 4

Now, we have 421.

Multiply 4 by 2 to get 8

Now, we have 8421.

Multiply 8 by 2 to get 6 as base-number and 1 as carry over.

Now, we have ₁68421.

Multiply base-number 6 by 2 to get 2 as base-number and 1 as carry over. Add the previous carry over 1 to get 3 as base number.

Now, we have ₁368421.

Carry on till you have a total of (19-1)/2 = 9 digits in place. And they would be 947368421. We call is Lower Half. To calculate the Upper Half, take 9's complement.

  999999999
 -947368421
------------
  052631578
------------

The result, therefore, is:

 0.052631578947368421

Enjoy! Now the example of 1/7. It will contain a total of 7 - 1 = 6 digits that will recur.

1/7 = 7/49. The number previous to 9 is 4 and one next to 4 is 5.

First, write 7. Multiply by 5, yielding 5 as base-number and 3 as carry overy - ₃57. The full chain is as below:

7 => 357 => 2857
  => 42857 => 142857

Therefore,

 1/7 = 0.142857

5. Trick - More Multiplication

This trick is an extension of the previous trick. It empowers you to multiply two numbers that can be expressed as z|y and z|(10-y).

The result of multiplication of the two numbers is z(z+1)|y(10-y)

To illustrate by an example,

37 * 33 = 3(3+1)|7*3 = 1221
141 * 149 = 14(15)|1*9 = 210|09 = 21009

Hmm... what do you say? Isn't maths really a fun!

4. Trick - Squaring numbers ending in 5

The sutra (formula) that empowers me to do so at ease is "Ekadhikena Purevena", meaning by one more than the previous one.

To calculate square of any number ending in 5, say z5 is z(z+1)|25, ie, multiply z by z+1 and attach 25 next to it.

For example, 75² is:

752 = 7(7+1)|25 = 5625

Simple and powerful, isn't it? Now, let us see what goes behind it. It's simple algebra.

(10x + 5)2 = 100x2 + 100x + 25
  = 100x(x+1) + 25
  = x(x+1)|25

3. Trick - Multiplication

Today we will learn about multiplication... once again, the same trick but with a difference.

The trick that I will tell you now will give you more power. After this, you will never need to know any table more than 5x5. I call this extension as "10's complement".

10's complement of any digit x is 10 - x. In two numbers that you are multiplying, take 10's complement of any digit greater than 5 and increase the digit next to it (higher significant value) by one. Reverse this process while calculating final value. The digits whose complements are taken shall be called "barred".

Calculation with barred digits will be considered as being done with negative numbers. For example, bar-4 * 4 = bar-16 and bar-8 + 3 = bar-5.

Now, I shall demonstrate it with an example:

    478*129 = 522*131

    522
   x131
 --------
   53342
   11
 --------

  = 62342 = 61662

2. Tip - Multiple Choice Questions

JEE (Joint Entrance Examination) consists of two phases: Screening, consisting of objective questions and Mains, consisting of subjective questions. Screening plays a major role in determining your entry - not only you need to be accurate, you also need to be fast.

In this article, let us see how can we avoid actual calculations and reduce time to arrive at the correct solution.

More often than not, you will not be required to completely solve the problem. You would be able to reject some answers by mere inspection. For example, 234*567 would be greater than 10,000. Now, 25*55 = 1375 (quick, mental calculation by Vedic Maths trick). So, the answer must be somewhere close to 137500 but less than it. How? Guess. Failed? Mail me!

Ok... that may be an artificial example. Let's take a real example - from 2004 screening.

Three distinct numbers are selected from first 100 natural
numbers. The probability that all the three numbers are
divisible by 2 and 3 is:

(a) 4/25            (b) 4/35
(c) 4/55            (d) 4/1155

According to requirement, the numbers must be divisible by 6. There are 16 numbers divisible 6. So, the probability is:

16C3 / 100C3

  16*15*14       16*15
----------- = ----------- (98/14 = 7)
 100*99*98     100*99* 7

       16
 < ---------- (1/99 < 1/96)
    100*6*7

       4         4
 < -------- < -------
    25*6*7     100*7

The only number that is less than 4/700 is 4/1155, hence the answer.

These calculations given above are very trivial, mere approximations. You never did an actual calculation. No calculation was complex. Well, except for 98/14 = 7, if you consider it.

1. Trick - Multiplication

Today we will learn about probably one of the first "Sutras" (meaning: formulas, Hindi) called Urdhva-Tiryak Sutra

In English, it means "Vertically and Crosswise". The demonstration below shows how one can shorten the calculation done:

Conventional Method	Vedic Method

How is it done? What's the trick? Well... nothing too great. It can be easily observed if you are very strong in Algebra. Ok.. you'll get it when I explain.

The Vedic calculation is done as told by the sutra: Vertically and crosswise. First, take unit digits of the two numbers, 2 and 6 in this case. Multiply. Units digit of the result, 2, is written in first row while tens digit, 1, is written in second row. The first row is result row. The second is the carry-over.

Now, take two digits at a time - 32 and 46 - multiply crosswise and vertically, ie, 3 by 6 and 2 by 8. Add. The result is 26. Add the carry-over. Units digit - 7 - is written in the result row. 2 is carry over.

Now, take three digits - 232 and 246. Multiple 2 by 6, 3 by 4 and 2 by 2. Add. Add the carryover - 2 - to it. You get 30.

Carry on till you are done will all digits - increasing the number of digits into consideration by one. One you reach the limit, start removing digits from the last.

For example, after you are done with 28232 and 53246, apply the same algorithm with 2823 and 5324. Notice that I've removed the units digits - 2 and 6. Now start removing one by one and complete.

Here is the algebraic explanation of the calculation:

          ax2 + bx + c
          dx2 + ex + f
   ----------------------------
      adx4 + (ae + bd)x3 + 
       (af + be + cd)x2 +
           (bf + ce)x + cf
   ----------------------------

Does it catch your eyes now? Still not... mail me at the email provided in contact info section.

Free-JEE Yahoo Group

Courtsey Rambo, an IITM alumnus, a Yahoo group has been created towards Free-JEE education. Click here to learn more about it.

Introduction

I am creating this blog with an intention to provide tips and tricks for all Indian Institute of Technology aspirants.

Given below is the list of all IITs (in alphabetical order):

• IIT Bombay
• IIT Delhi
• IIT Guwahati
• IIT Kanpur
• IIT Kharagpur
• IIT Madras
• IIT Roorkee

Below is the list of the website of their alumni networks:

• IIT Bombay
• IIT Delhi
• IIT Guwahati
• IIT Kanpur
• IIT Kharagpur
• IIT Madras
• IIT Roorkee