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 Post subject: Number Theory Challenge 2Posted: Wed, 4 Aug 2010 19:26:10 UTC
Prove that, if n is an arbitrary positive integer then the expression

is divisible by 5 iff n is not divisible by 4.

enjoy,

cheers

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 Post subject: Re: Number Theory Challenge 2Posted: Thu, 5 Aug 2010 06:34:46 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 7624
Location: NCTS/TPE, Taiwan
Casdan1 wrote:
Prove that, if n is an arbitrary positive integer then the expression

is divisible by 5 iff n is not divisible by 4.

enjoy,

cheers

Spoiler:
Power sum in ...

It follows immediately from the existence of primitive root: , so when we must have , and obviously when we have .

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 Post subject: Re: Number Theory Challenge 2Posted: Thu, 5 Aug 2010 16:16:24 UTC
outermeasure wrote:
Casdan1 wrote:
Prove that, if n is an arbitrary positive integer then the expression

is divisible by 5 iff n is not divisible by 4.

enjoy,

cheers

Spoiler:
Power sum in ...

It follows immediately from the existence of primitive root: , so when we must have , and obviously when we have .

Yep you got it.

Below is your solution translated into plain english.

Each of the numbers is of the form , where k is an integer.

Also note that if A stands for 1,2,3, or 4 and if l is a positive integer, then is of the form because where the first factor on the right is divisible by 5, and the second is an integer.

Any positive integer may be written in the form where l is a positive integer or zero, and the remainder r is one of the numbers 0, 1, 2, or 3. In particular the exponent n may be written in this form so that.

where m is an integer, and .

Thus, S is divisible by 5 iff R is. If n is divisible by 4, then r = 0 and R = 4; in this case S is not divisible by 5. But if n is not divisible by 4, then r = 1, 2, or 3, and the corresponding values of R are 10, 30 or 100; in this case S is divisible by 5.

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 Post subject: Re: Number Theory Challenge 2Posted: Thu, 5 Aug 2010 21:51:16 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 15563
Location: Austin, TX
Casdan1 wrote:
outermeasure wrote:
Casdan1 wrote:
Prove that, if n is an arbitrary positive integer then the expression

is divisible by 5 iff n is not divisible by 4.

enjoy,

cheers

Spoiler:
Power sum in ...

It follows immediately from the existence of primitive root: , so when we must have , and obviously when we have .

Yep you got it.

Below is your solution translated into plain english.

Each of the numbers is of the form , where k is an integer.

Also note that if A stands for 1,2,3, or 4 and if l is a positive integer, then is of the form because where the first factor on the right is divisible by 5, and the second is an integer.

Any positive integer may be written in the form where l is a positive integer or zero, and the remainder r is one of the numbers 0, 1, 2, or 3. In particular the exponent n may be written in this form so that.

where m is an integer, and .

Thus, S is divisible by 5 iff R is. If n is divisible by 4, then r = 0 and R = 4; in this case S is not divisible by 5. But if n is not divisible by 4, then r = 1, 2, or 3, and the corresponding values of R are 10, 30 or 100; in this case S is divisible by 5.

I like outermeasure's way better, it's more transparent.

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 Post subject: Re: Number Theory Challenge 2Posted: Fri, 6 Aug 2010 15:02:17 UTC
Casdan1 wrote:
outermeasure wrote:
Casdan1 wrote:
Prove that, if n is an arbitrary positive integer then the expression

is divisible by 5 iff n is not divisible by 4.

enjoy,

cheers

Spoiler:
Power sum in ...

It follows immediately from the existence of primitive root: , so when we must have , and obviously when we have .

Yep you got it.

Below is your solution translated into plain english.

Each of the numbers is of the form , where k is an integer.

Also note that if A stands for 1,2,3, or 4 and if l is a positive integer, then is of the form because where the first factor on the right is divisible by 5, and the second is an integer.

Any positive integer may be written in the form where l is a positive integer or zero, and the remainder r is one of the numbers 0, 1, 2, or 3. In particular the exponent n may be written in this form so that.

where m is an integer, and .

Thus, S is divisible by 5 iff R is. If n is divisible by 4, then r = 0 and R = 4; in this case S is not divisible by 5. But if n is not divisible by 4, then r = 1, 2, or 3, and the corresponding values of R are 10, 30 or 100; in this case S is divisible by 5.

I like outermeasure's way better, it's more transparent.

Yeah, his solution is very elegant and compact. However, I'm completely lost just by some of his notation and terminology.

I'm a Physicist in training (Grad School), so for me Math is just a tool to use to get what I want.
One day we will be able to peacefully coexist, (Physicists vs. Mathematicians).

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 Post subject: Re: Number Theory Challenge 2Posted: Fri, 6 Aug 2010 21:34:11 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 15563
Location: Austin, TX
Casdan1 wrote:
Yeah, his solution is very elegant and compact. However, I'm completely lost just by some of his notation and terminology.

I'm a Physicist in training (Grad School), so for me Math is just a tool to use to get what I want.
One day we will be able to peacefully coexist, (Physicists vs. Mathematicians).

That's all well and good, but all I said was it was more transparent.

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 Post subject: Re: Number Theory Challenge 2Posted: Sat, 7 Aug 2010 04:22:23 UTC
 Moderator

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 7624
Location: NCTS/TPE, Taiwan
Casdan1 wrote:
Below is your solution translated into plain english.

No! That was not my solution.

Spoiler:
Working mod p, {1,2,...,(p-1)} and {g,2g,...,(p-1)g} both form a reduced residue system, where g, as is customary, denotes a primitive root. So

and so

Now since g is a primitive root, we know if and only if , and if we have . So

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