As explained in earlier posts, osculation is the process of multiplying the last one or more digits of a number by some constant and adding the result to the left-over part of the number (positive osculation) or subtracting the result from the left-over part of the number (negative osculation). If the osculation involves just the last digit of the original number, it is called simple osculation. If the osculation involves more than one digit, it is called multiplex osculation. We also saw in the post on divisibility tests how to use osculation in divisibility tests for numbers like 7, 13, etc., even though we did not, at that time, know that we were using osculation.

The question naturally arises as to why or how osculation works in determining divisibility. In this post, I will look at the mathematics of how osculation enables us to test for divisibility. To start on this path, we need to understand the concept and properties of a mathematical operator known as the modulus operator. Most of us are probably already familiar with this operator, but we will use the first part of this post to re-examine the modulus operator, and then we will look at its relationship to osculation.

The modulus operator, to put it in simple terms, gives us the remainder of the division of the first operand by the second operand. It is usually denoted by the term “mod” between the first and second operands (in many programming languages, the modulus operator is implemented as a function that takes the two operands as arguments). Usually, the result of the modulus operator is defined only when the second operand is positive. Some examples of the results from the modulus operator are given below.

- 25 mod 5 = 0
- 26 mod 4 = 2
- 31 mod 2 = 1
- 12568 mod 11 = 5
- 3 mod 28 = 3
- 7 mod 7 = 0
- -31 mod 6 = 5 (-36 is the highest multiple of 6 below –31, and the difference between –36 and –31 is 5)

Here are some properties of the modulus operator that we can readily derive from its definition. We will restrict ourselves to the use of the modulus operator when the second operand is positive.

- The results of the modulus operator can range from 0 to n – 1, where n is the value of the second operand
- x mod y = x when x < y
- x mod y = 0 when x = y
- if x mod y = 0, then x is divisible by y
- if x mod y = 0, then ax mod y is also 0, where a is any constant. This means that if y is a factor of x, then y, by necessity, has to be a factor of ax also
- But the reverse is not true: thus if ax mod y = 0, that does not mean that x mod y = 0. As an example, 2*4 mod 8 = 0, but 4 mod 8 is not equal to 0
- In general, if a and y are not co-prime, then it is possible for ax mod y to be 0 while x mod y is not. If a and y are co-prime, then if ax mod y = 0, then x mod y = 0
- Regardless of the value of x mod y, yx mod y is always 0 (multiplying the first operand by the second operand makes the second operand a factor of the product)
- if x mod y = a, and z mod y = b, then (x + z) mod y = (a + b) mod y
- x mod y = (x + ay) mod y (this follows directly from the previous property since ay mod y is always 0)
- if x mod y = a, and z mod y = b, then (x – z) mod y = (a – b) mod y
- the previous rule then leads directly to this one: x mod y = (x – ay) mod y
- x mod y + (-x) mod y = y if x is not divisible by y
- x mod y + (-x) mod y = 0 if x is divisible by y. Combined with the third rule above, this implies that (-x) mod y is also equal to 0 if x is divisible by y

Now, let us look at a divisibility rule that uses osculation, in the context of the properties above to try and understand how it works. One of the simplest such rules is that if the positive osculation of a number by 5 is divisible by 7, then the original number is divisible by 7. Thus, for instance, 364 is divisible by 7 because 36 + 5*4 = 56, which is divisible by 7.

To understand why this works, let us first break the given number into two parts. The first part will take into account all the digits except the last digit, and the second part will take into account the last digit. Thus, 364 = 36*10 + 4. In general, any number with more than one digit can be rewritten as 10x + y, where x and y are non-negative integers.

What does the number (10 x + y) being divisible by 7 mean?

It means that (10x + y) mod 7 = 0.

Since x mod y = 0 implies that ax mod y = 0, let us multiply the first operand by 5 (note that 5 and 7 are co-primes, so 5x mod 7 = 0 implies also that x mod 7 = 0). We get 5*(10x + y) mod 7 = 0.

Expanding this out, we can then say that (50x + 5y) mod 7 = 0.

Now, recognizing that 49x is a multiple of 7 (since 49 is divisible by 7, and x is some non-negative integer), we can subtract 49x from the first operand without affecting the result of the modulus operator. Thus, (x + 5y) mod 7 = 0.

But, that is precisely what the divisibility rule says. Recognize that x + 5y is the positive osculation of (10x + y) by 5. Thus, step by step, we have derived the positive osculation divisibility rule for 7.

We can do the same thing for the negative osculation test for divisibility by 7 also. Remember that the test says that if the negative osculation of a number by 2 is divisible by 7, then the number is divisible by 7. Thus, in the case of 364, it is divisible by 7 because 36 – 2*4 = 28 is divisible by 7.

Once again, consider the given number to be 10x + y. 10x + y being divisible by 7 implies that (10x + y) mod 7 = 0.

Multiply the first operand by 2. This gives us (20x + 2y) mod 7 = 0 (once again, note that since 2 and 7 are co-prime, 2x mod 7 = 0 implies that x mod 7 = 0).

Recognize that 21x is a multiple of 7 because 21 is divisible by 7 and x is a non-negative integer. We can then subtract 21x from the first operand without affecting the result of the modulus operator. This gives us (-x + 2y) mod 7 = 0.

Since x mod y = (-x) mod y when x is a multiple of y, we can multiply the first operand by –1 to get (x – 2y) mod 7 = 0.

Thus, (x – 2y) mod 7 = 0 implies that (10x + y) mod 7 = 0. This proves the negative osculation divisibility test for 7.

As you can guess at this point, the practitioners of Vedic Mathematics simply codified all this into the concept of natural osculators or vestanas. It is easy to identify multiples of a number that end in 9 or 1 like we did above for 7 (49 and 21) so that we can reduce the 10x in (10x + y) to just x by subtracting the multiple of the number that ends in 9 or 1.

But, now that we know the secret behind osculation and its use of the properties of the modulus operator to test for divisibility, we can derive our own divisibility tests without any problems. And some of them could be much better or more convenient to use than the tests we learned about in the post on divisibility rules.

For example, let us put to good use the facts that 1001 is a multiple of 7, and any number can be expressed as 1000x + y (with non-negative x and y). Obviously, this works best when the number is at least 4 digits long.

(1000x + y) mod 7 = 0 implies that (-x + y) mod 7 = 0 (we simply subtracted 1001*x, which is a multiple of 7 from the first operand).

This then tells us that (x – y) mod 7 = 0.

Let us test this out on 27,118. In this case, x = 27 and y = 118. x – y = –91, which is divisible by 7. We can verify that 27118 is also divisible by 7. We just derived another divisibility rule for 7! This divisibility rule uses multiplex osculation rather than simple osculation.

Similarly, 1001 is a multiple of 13 too. So, we can say that if the number can be expressed as 1000x + y, and x – y is divisible by 13, then the number is divisible by 13. Take the number 50,037, for instance. Since 50 – 37 = 13 is divisible by 13, we can say that 50,037 is divisible by 13. It is easy to verify that this is indeed the case. With very little effort, we have derived another divisibility rule for 13 based on multiplex osculation!

Obviously, having a multiple that is one away from a power of 10 makes this convenient, but the principle works even if that is not the case. For instance, 1003 is a multiple of 17. Thus, if a number can be expressed as 1000x + y, then (1000x + y) mod 17 = 0 implies that (–3x + y) mod 17 = 0 (we just subtracted 1003x, which is a multiple of 17 from the first operand). The result is not a divisibility rule based on osculation, but might be just as convenient to use.

For instance, take the number 83,215. In this case, x = 83 and y = 215. y – 3x = 215 – 249 = –34. Since –34 is divisible by 17, we can conclude that 83,215 is also divisible by 17. It is easy to verify that this is indeed true.

992 is a multiple of 31. Using that fact, we can derive the divisibility rule that if the number can be expressed as (1000x + y), then if (8x + y) is divisible by 31, then the original number is divisible by 31 too. Thus 3007 is divisible by 31 because x = 3 and y = 7, and 8x + y = 24 + 7 = 31, which is obviously divisible by 31.

Now that we have unraveled the secrets of osculation and how the combination of osculation and the modulus operator results in divisibility tests, we can all go wild deriving dozens of divisibility tests for any number we are interested in. You might discover divisibility rules that are much easier to apply or appeal to you for some other reason. All you have to do is find a multiple of the divisor that is in the vicinity of a power of 10, and you are all set! Try a few. I am sure you will be hooked on it. Discovery is a powerful and addictive feeling!!

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