Hello Everyone

Can anyone explain why we cannot use 'osculators' to check divisibility for an even number or a number ending with 5?

Results 1 to 4 of 4

- Nov 3rd 2010, 03:16 PM #1

- Joined
- Nov 2010
- Posts
- 2

- Nov 4th 2010, 02:47 PM #2

- Joined
- Apr 2005
- Posts
- 19,336
- Thanks
- 2857

- Nov 4th 2010, 05:22 PM #3

- Joined
- Nov 2010
- Posts
- 2

Ok, I found the answer in another site. Just copy-pasting the info here for informational purpose.

Osculator Method/Seed of a number

Osculator/seed of a number can be negative and positive

Let the number be n,then if nk+1=10m then positive osculator of n is m

And if nk-1=10m then the negative osculator of n is m

Thus if the number is 7

We have 7(3)=21 and 21-1=10(2) so the negative osculator of 7 is 2

We have 7(7)=49 and 49+1=10(5) so the positive osculator of 7 is 5

This osculator is used to check for the divisibility by 7 , 13 and other prime numbers for which there are no straight forward rules.

Osculator for an even number or a number ending with 5

If the number is even then nk+1 and nk-1 are odd which cant be divisible by 10,hence the osculator of an even number doesn’t exist

Similarly when a number ends with 5,then also the osculator is not possible

Thus only odd numbers have their osculators and which aren’t also multiples of 5.

Examples :

1) Lets see whether 7896834 is divisible by 7 or not?

The negative osculator of 7 is 2

We will do the following operations on the number

789683 -4(2)=789675

78967-5(2)=78957

7895-7(2)=7881

788-1(2)=786

78-6(2)=66

6-6(2)=-6 which is not divisible by 7

Hence the number is NOT divisible by 7

Infact the thing which we are doing is this

7896834mod7=>(7.10^6 + 8.10^5 + 9.10^4 + 6.10^3 + 8.10^2 + 3.10^1 + 4.10^0 mod7)

Now gcd(7,10)=1

So we can find integers r and s such that 7r+10s=1

We know 7(3)-1=10(2)=> the negative osculator of 7 is 2

7(3)+10(-2)=1 hence the value of s=-2

Then we can say that

(7.10^6 + 8.10^5 + 9.10^4 + 6.10^3 + 8.10^2 + 3.10^1 + 4.10^0 mod7)

=>(7.(-2)^6 + 8.(-2)^5 + 9.(-2)^4 + 6.(-2)^3 + 8.(-2)^2 + 3.(-2)^1 + 4.(-2)^0 mod7)

2) Lets see whether 49140 is divisible 13 or not?

The positive osculator of 13 is 4 (coz 13.3+1=10(4))

We will do the following operations on the number

4914+0(4)=4914

491+4(4)=507

50+7(4)=78

7+8(4)=39

Now since 39 is divisible by 13 so is the number itself

- Nov 5th 2010, 02:59 AM #4

- Joined
- Aug 2009
- Posts
- 170
- Thanks
- 8

It's from the definition nk+1=10 or nk-1=10

The reason is because even numbers and 5 dont have osculators ... it's right there in the text you copied

if n=2, then you'll note that nk+1 and nk-1 will always be odd, so they can't be divided by 10.

Similar case for n=5

Click on a term to search for related topics.