Omnimaga

General Discussion => Other Discussions => Math and Science => Topic started by: Munchor on March 18, 2011, 05:46:18 pm

Title: Why is this?
Post by: Munchor on March 18, 2011, 05:46:18 pm
x^n

if last digit of x = 9 and n is even:

the last digit of n is 1.


Example

9^4 = 6561
69^2 = 4761

Any number

Can any of you explain me why?
Title: Re: Why is this?
Post by: Ashbad on March 18, 2011, 05:50:09 pm
9^2 = 81
9^4 = 6561
9^6 = 531441
9^8 = 43046721
9^10 = 3486784401

And...

19^2 = 361
29^2 = 841
39^2 = 1521
49^2 = 2401
59^2 = 3281

... if it helps prove scout's legit point.
Title: Re: Why is this?
Post by: merthsoft on March 18, 2011, 05:50:33 pm
Because 9*9 = 81, and when you keep multiplying by 9, the last number will bounce between 9 and 1. So when the power is odd, it will be 9, when it's even it'll be 1.
Title: Re: Why is this?
Post by: leafy on March 18, 2011, 05:54:21 pm
9^(2n) mod 10 = 1
and 9^(2n-1) mod 10 = 9.

It's the same thing with 2 - 2, 4, 8, 6, 2, 4, 8, 6, etc.
Title: Re: Why is this?
Post by: jnesselr on March 18, 2011, 09:46:05 pm
Very nice and simple proof.  How about a slightly harder one.  Why can you check divisibility by 3 or 9 by adding up all the digits of a number?
Title: Re: Why is this?
Post by: Munchor on March 19, 2011, 08:07:04 am
Quote from: graphmastur on March 18, 2011, 09:46:05 pm
Very nice and simple proof.  How about a slightly harder one.  Why can you check divisibility by 3 or 9 by adding up all the digits of a number?

That's harder.

Quote from: leafiness0 on March 18, 2011, 05:54:21 pm
9^(2n) mod 10 = 1
and 9^(2n-1) mod 10 = 9.

It's the same thing with 2 - 2, 4, 8, 6, 2, 4, 8, 6, etc.

Quote from: merthsoft on March 18, 2011, 05:50:33 pm
Because 9*9 = 81, and when you keep multiplying by 9, the last number will bounce between 9 and 1. So when the power is odd, it will be 9, when it's even it'll be 1.

Thanks!
Title: Re: Why is this?
Post by: calcdude84se on March 19, 2011, 04:54:09 pm
I'll take it :)
So, a base-10 number can be represented as an*10n+an-1*10n-1+...+a1*10+a0
You can find divisibility by nine by finding the remainder when the number is divided by nine, in other words, by taking the number mod 9.
So, 10 = 1 (mod 9), and also 10n = 1 (mod 9)
This means that an*10n+an-1*10n-1+...+a1*10+a0 = an+an-1+...+a1+a0 (mod 9)
So, the remainder of a number divided by 9 is equal to the remainder of the sum of its digits written in base 10 by 9.
More generally, this is the divisibility test for n-1 in base n. :D
Title: Re: Why is this?
Post by: AngelFish on March 19, 2011, 05:40:20 pm
That only addresses (A mod 9 = 0) iff (0 = <digital sum> mod 9). He also asked about (A mod 9 = 0) iff (0 = <digital sum> mod 3).

Since it's been demonstrated that the test holds for 9 and 3 is a prime factor of 9, the test holds for some subset of the set of all multiples of 3. However, it also appears to hold for all multiples of 3, which that test alone cannot account for. Well, actually it can. The proof is essentially that used for 9. But I had fun testing all integers up to 12,000 with my BASIC program  :P
Title: Re: Why is this?
Post by: willrandship on March 19, 2011, 05:41:48 pm
BTW, the first post can be a little misread. You should define it with another letter. n cannot be even and start with one. :P
Title: Re: Why is this?
Post by: AngelFish on March 19, 2011, 05:54:15 pm
Okay, finished testing all multiples of 3 through 12,000 against the digital sum test. It works without exceptions :D

Sometimes knowing the answer just isn't as satisfying as brute forcing it.
Title: Re: Why is this?
Post by: jnesselr on March 20, 2011, 01:42:12 pm
lol, yeah.  good job qwerty.55.