Special Divisibility rules | one shortcut method for all numbers including primes.

Sohel Sahoo: Hello guys,

In this article you would come to know an amazing formula to check divisibility rule of any number that could help you a lot to solve problems in hands down in exams.

Read Also: Divisibility Rules formation from 2to19

Rules for applying the formula:-

There exists two specific no.s (called osculators,viz:-one is positive & other being negative) for each number.

1. Firstly, multiply the osculator to the unit digit of the given number, Which is to be examined whether it is divisible by that divisor no. Or not.

2. Then add the result so obtained on multiplication from the rest of the digits gained by cancelling that unit digit of the given number.

3. Now,if the result of this sum is that very prime no. Or a multiple of it then the given number is said to be divisible by the divisor number.

We can carry on this process repeatedly until we reach a comparatively small no. Which gives us the necessary clue as to whether the given no. is divisible by it or not.

Note:- In case of positive osculator we can directly follow the above said procedure while for negative osculator the exception is addition with it becomes subtraction from rest digits being multiplied its modulus (+ve) value.

Procedures to find the osculators:-

1. Let the osculator be 'x'.
Hence, according to rules cited above multiply it to last digit of given number.Express it as the sum of the result of multiplication and rest digits.

2. Now check for which value of 'x' the expression is divisible by the divisor number.

You can check from 1,2,3... So on for positive osculator manually and -1,-2,-3 ...so on for negative osculator.

Note:- The divisor number should be of two digits if not then express it to the nearer multiple of two digits as in case of one digits its previous digits are zero which is of no use !

Relationship between two osculators:-

Let the given no.which is to be tested be represented as ' ab' & osculator be ' x'.

So,according to formula we have bx+a, divisible by ab.

Now, bx+a—ab² is divisible by ab

=>a—ab²+bx is divisible by ab

=>a—b(ab--x) is divisible by ab

^{Clearly,(ab--x) is the negative osculator.}

NOTE:- (i) Sum of both osculators (positive&negative) is equal to divisor.

^{[Excluding the sign (+)/(--)]}

^{(ii) By getting one we can obtain another just by placing opposite sign to difference of that osculator from divisor.}

^{(iii)For easy & efficacy take that osculator which is small irrespectve of signs.}

e.g.-

# Find the osculator for 13.

Let the osculator be 'x'.
According to rules, the required expression will be '3x+1'.
Now checking the values of x for which the 3x+1 is divisible by 13 we find
3×4+1=13; divisible by 13.
Hence, the osculator will be '4' for 13.
Also,3×(-9)+1=-26; divisible by 13.
Negative osculator for 13 is '-9' .

# Similarly,if we try to find osculator for 7 (being single digit)then primarily we have to convert it to nearer multiple i.e.14.
Now, Check value of'x'for which 4x+1 is divisible by 7.
We find,4×5+1=21; divisible by 7
Hence, osculator for 7 is '5'.
Also 4×(-2)+1=-7; divisible by 7.
Hence, negative osculator for 7 is '-2'.

Likewise, you can test osculators for any divisor number which is generalized.

***Shortcut pro* tips for finding osculators:-***

Any divisor no. ending in 9 can generally be written as 'a9'.
Let the osculator be 'x'.
According to rules the expression will be '9x+a'.
In decimal representation we have;
a9=10a+9
=9a+a+9
=9a+9+a
=9(a+1)+a
~9x+a
Obviously, the positive osculator is (a+1).
# Henceforth, owing to this the positive osculator for divisor numbers ending in 9 is just one more than its previous digits.

1. For 9,19,29,39 etc.(all ending in 9);the positive osculators are 1,2,3,4 etc.

2. For 3,13,23,33 etc. (all ending in 3) multiply them by 3 in order to get 9 in 1's place as 9,39,69,99 etc.Thus you get 1,4,7,10 etc.as positive osculators.

3. For 7,17,27,37 etc.(all ending in 7) multiply them by 7 so as to attain 9 in unit place like 49,119,189,259 etc. Hence,you get 5,12,19,26 etc.as positive osculators.

4. For 1,11,21,31 etc. (all ending in 1) multiply them by 9 thereof attain 9 as last digit as 9,99,189,279 etc. So, you get positive osculators viz. 1,10,19,28 etc.

For the sake of your understanding I have given osculators for a set of no.s to visualise &feel comfortable in this concept as reference.

Number	Positive Osculator	Negative Osculator
3	1	-2
7	5	-2
9	1	-8
11	10	-1
13	4	-9
17	12	-5
19	2	-17
21	19	-2
23	7	-16
27	19	-8
29	3	-26
31	28	-3
33	10	-23
37	26	-11
39	4	-35
41	37	-4
43	13	-30
47	33	-14
49	5	-44
51	46	-5
53	16	-37
57	40	-17
59	6	-53
61	55	-6
63	19	-44
67	47	-20
69	7	-62
71	64	-7
73	22	-51
77	54	-23
79	8	-71
81	73	-8
83	25	-58
87	61	-26
89	9	-80

1. Test whether 416 is divisible by 13 or not.

Sol:- The suitable osculator for 13 is 4.
So, as cited above check 6 x 4+41=65 if divisible by 13
Clearly it do so , that's why 416 is also divisible by 13.

2.Test whether 69125 is divisible by 7 or not.

Sol:- For 7 suitable osculator will be -2.
So, check 6912--(5 x 2)=6902 if divisible by 7.
Proceeding the said process repeatedly then check 690--(2x2)=686 if divisible by 7.
Then,68--(2x6)=56 is divisible by 7 .So, the no. 69125 is divisible by 7 aswell.

You can check this process by taking 5 also as follows.

Here 21 is divisible by 7;that's why 69125 is also divisible by 7.