Search The Web

Today's Headlines

Thursday, October 22, 2009

Vedic Mathematics Lesson 25: Multiplex Osculation

In the last lesson, we learnt about simple osculation and its utility in divisibility tests. The "simplicity" of simple osculation comes from the fact that we osculate one digit at a time. When we take the remainder after division by 10, and the integer quotient of division by 10, we separate the last digit from the rest of the number, and then perform osculation with it.

You can find all the previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2

Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I
Straight Division II
Vinculums
Divisibility Rules
Simple Osculation

In this lesson, we will extend the method to osculation by more than one digit. This is equivalent to taking the remainder after division by a higher power of 10, multiplying it by some number and adding this result to, or subtracting it from, the integer quotient of division by the higher power of 10. This is called multiplex osculation.

Let us illustrate this procedure by trying it out on a few numbers.

For instance, supposed we have to positively osculate 487537 by 10, 2 digits at a time. The procedure simply consists of taking the last 2 digits, multiplying them by 10, and adding them to the number left over after removal of the last 2 digits. Thus, the given osculation would be performed as 4875 + 37x10 = 4875 + 370 = 5245. Performing the same osculation once more would give us 52 + 45x10 = 502. One more iteration would lead us to 5 + 02x10 = 25.

As a second example, we will try negatively osculating 4378374 by 2, 3 digits at a time. The procedure consists of taking the last 3 digits, multiply them by 2, and subtracting the result from the number left over after removal of the last 3 digits. Thus, the given osculation would be performed as 4378 - 374x2 = 4378 - 748 = 3630. When we perform the same osculation once more, we get 3 - 630 x 2 = 3 - 1260 = -1257.

Now that we know how to perform multiplex osculation, let us see why the procedure is useful.

In the previous lesson, we figured out how to derive the natural positive and negative simple osculators for any number that ends in 1, 3, 7 or 9. Unfortunately, we also figured out that this procedure can lead to large simple osculators that can make the test for divisibility a little difficult.

Multiplex osculators can make that burden a little lighter under some circumstances. That is why it is important to learn multiplex osculation. However, just so I keep expectations low, testing divisibility by large numbers is always going to be somewhat difficult, and multiplex osculation is not going to make it trivial by any stretch of the imagination.

Let us deal with some simple cases first and then see how to convert other cases to these simple cases if possible.

The first set of simple cases are numbers that end with a series of 9's. For instance, take the number 399. We can recognize immediately that its natural positive simple osculator is 40 (one more than the number left after dropping the 9 at the end of the number). However, osculation by 40, one digit at a time (simple osculation by 40) might get a little tedious. Vedic mathematics says that we don't have to do that: we can simple take one more than the number left over after removing the series of 9's and osculate with that number over as many digits as the number of 9's in the series.

To put it in practical terms, in the case of 399, the number left over after removing the series of 9's is 4. The number of 9's in the series is 2. Thus, we could say that P2 = 4. What the notation means is that for 399, the multiplex positive osculator for osculation by 2 digits at a time is 4. To test this, let us take a number like 1596 and check its divisibility by 399 using multiplex osculation. Since P2 = 4 for 399, we perform positive multiplex osculation of 1596 by 4. Our first iteration gives us 15 + 96 x 4 = 399. We see that 399 is divisible by 399, hence we conclude that 1596 is divisible by 399. We can verify by performing the actual division that this is indeed the case.

Applying the same logic to a few other examples, we can derive the natural positive multiplex osculators for the numbers below:

499: P2 = 5
2999: P3 = 3
1199: P2 = 12
999: P3 = 1
79999: P4 = 8

The second set of simple cases are numbers that end in a 1 preceded by a series of 0's. Take the number 2001, for instance. We can recognize immediately that is natural negative simple osculator is 200 (the number left over after dropping the 1 at the end of the number). However, osculation by 200 one digit at a time (simple osculation by 2) may sound somewhat inconvenient. Instead, Vedic mathematics provides us with following rule: The natural negative multiplex osculator of the number is the number left after dropping the series of 0's and the final 1. The number of digits to osculate by is the number of digits dropped.

To put it in practical terms, in the case of 2001, the number left after dropping the series of 0's and the final 1 is 2. The number of digits dropped is 3. Thus, we say that N3 = 2. The notation means that the negative multiplex osculator is 2 and the number of digits to osculate by is 3. Suppose we have to test the divisibility of 130065 by 2001. We perform negative osculation by 2, 3 digits at a time to get 130 - 065x2 = 0. Since 0 is divisible by 2001, we conclude that 130065 is divisible by 2001. Indeed it is easy to verify that 130065 is divisible by 2001.

Applying the same logic to a few other examples, we can derive the natural negative multiplex osculators for the numbers below:

21001: N3 = 21
501: N2 = 5
8001: N3 = 8
1501: N2 = 15
20001: N4 = 2

The natural multiplex osculators for other numbers (that do not end with a series of 9's or a 1 preceded by a series of 0's) can be derived if those numbers can be converted into numbers of these two types by multiplication with some other number. As you can imagine, this is not a completely straightforward process that can be done using a standard set of rules.

The easiest way to do this is to become familiar with the repeating decimals that result from division by certain numbers. For instance, 1/7 = 0.142857 repeated continuously. It is also easy to see that 142857 x 7 = 999999. Thus, we can say that P6 for 142857 is 1.

Similarly, 1/11 = .09, repeated continuously. Thus, it is easy to see that 90909 x 11 = 999999. Thus, we can say that P6 for 90909 is 1.

One might also be able to guess that 143 x 7 = 1001, so N3 for 143 is 1.

This is not a surefire method and will not work for many numbers. For many numbers, a little trial and error is necessary to find a multiplier that will convert the number to something for which the multiplex osculator is readily available. The idea is to get as many 9's at the end as possible so that the positive osculator becomes smaller. Ultimately, if one is able to find a multiplier that creates 28999, it is better than a multiplier that creates 7219. Similarly, a multiplier that creates 54001 is better than one that creates 27211.

Moreover, note that a multiplex osculator that osculates over too many digits is not particularly useful for testing divisibility. Take the case of the number 2857, for instance. After some trial and error, we may be able to figure out that 2857 x 7 = 19999. Thus we can say that P4 for 2857 is 2. But, let us see if we can apply this method to test divisibility of 8571 by 2857.

Applying the osculation rule, we immediately see that the results of the multiplex osculation are even worse than the original number when it comes to ascertaining divisibility by 2857. Successive multiplex osculations of 8571 by 2 give us 17142, 14285, 8571, 17142, 14285, 8571, etc. repeating continuously. Thus, multiplex osculation is not as useful as we found simple osculation to be.

In spite of the disadvantages associated with multiplex osculation, it is important to know the procedure for the sake of completeness. Moreover, there are some cases where multiplex osculation can give us answers to divisibility questions faster than simple osculation can. Practicing both forms of osculation is key to understanding when to use which method and what the advantages and disadvantages of each method are. Hope you take the time to practice so that you are expert at both and can choose whichever method gives you the best advantage, given the circumstances. Good luck, and happy computing!

1 comment:

The Vedic Maths Forum India said...

Hello there!
We Like your posts on Vedic Maths. We are from The Vedic maths Forum India and were wondering if you would like to contribute to our blog?

http://vedicmathsindia.blogspot.com

You could mail me on gtekriwal(at)vedicmathsindia(dot)org

If you agree I could then add you as our Blog Author and will get u introduced to our readers.

Let us know.

Thanking you
Gaurav Tekriwal
President
The Vedic Maths Forum India

Visitors Country Map

Free counters!

Content From TheFreeDictionary.com

In the News

Article of the Day

This Day in History

Today's Birthday

Quote of the Day

Word of the Day

Match Up
Match each word in the left column with its synonym on the right. When finished, click Answer to see the results. Good luck!

 

Hangman

Spelling Bee
difficulty level:
score: -
please wait...
 
spell the word:

Search The Web