Omnimaga

General Discussion => Other Discussions => Math and Science => Topic started by: Anima on February 24, 2014, 06:35:11 am

Title: Terminating decimal
Post by: Anima on February 24, 2014, 06:35:11 am

Hey, I had a math exam today and there was the following problem: For which positive integers n is (4n + 1)/(n * (2n - 1)) a terminating decimal?
With terminating decimals are meant non-periodic numbers like 1.25 or -2.22, I guess. My first idea was to write this fraction as 6/(2n - 1) - 1/n or 4/(2n - 1) + 1/(2n^2 - n) and then create a system of equations, where the first and the second denominator are equal to something like 2^p * 5^q.
How would you proof this?

Title: Re: Terminating decimal
Post by: Xeda112358 on February 24, 2014, 07:05:30 am

You only need to worry about when n(2n-1) is not of the form 2ⁿ5^m.

2n-1 is always odd, so that must always be a power of 5.

2n-1=5^k ==> n=(5^k-1)/2

So then you need (5^k-1)/2 to be factored as powers of 2 and 5. 5 will never be a factor, so we need (5^k-1)/2=2^m ==> 5^k-1=2^m+1

This is true for k=1, which is n=2.

All other cases are when (4N+1) happens to cancel any non-2 and non-5 factors from the denominator, which happens at N=8. (2n-1 = 15, 4n+1=33, so the 3 cancels leaving 2^3*5 in the denominator).

Title: Re: Terminating decimal
Post by: Anima on February 24, 2014, 07:46:23 am

Thank you :)

Quote from: Xeda112358 on February 24, 2014, 07:05:30 am

2n-1=5^k ==> n=(5^k-1)/2

But this should be n = (5^k+1)/2, right?

Title: Re: Terminating decimal
Post by: Xeda112358 on February 24, 2014, 07:47:53 am

Oops, yeah.

Title: Re: Terminating decimal
Post by: Anima on February 24, 2014, 08:16:44 am

But this would mean that 2 will never be a factor of (5^k+1)/2, because it is always odd. So, (5^k+1)/2=5^m?

Title: Re: Terminating decimal
Post by: Xeda112358 on February 24, 2014, 09:57:23 am

Yep, that is correct. 5^k+1 mod 100 = 26, so  (5^k+1)/2 mod 10 = 3. Then (5^k+1)/2=5^m is never true, so the only way to take care of that is if (5^k+1)/2=(4n+1)2^r, but since it is odd, we know it must be that r=0, so (5^k+1)/2=(4n+1)

Also, sorry if I am wrong about anything, I haven't slept much.