Problem 44 - Page 2

Jameson · #16 $Old$ February 1st, 2008, 11:40 PM

Quote:

Originally Posted by Skinner $View Post$

oh. I messed up. Fail.

Good try though! Don't be discouraged. I have failed much worser doing far easier proofs.

Aryth · #17 $Old$ February 6th, 2008, 02:43 PM

The Sum:

$\sum_{i = 1}^{k}\frac{1}{k}$

Does not yield an integer for all $k \geq 2$ .

For k = 2:

$1 + \frac{1}{2} = \frac{3}{2}$

That is not an integer.

Since the series is divergent, then we must reduce it:

$\left[\frac{(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1)}{(2*3*...*k)}\right]$

To have an integer it has to be that for some integer M:

$\left[\frac{(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1)}{(2*3*...*k)}\right] = M$

$(3*4*...*k) + (2*4*...*k) + ... + (2*3*...*k-1) = (2*3*...*k)*M$

This can be rewritten as:

$k*((k-1)*((k-2)*...(2+3) + 2*3) + 2*3*4)...) + (2*3*...*k-1) = (2*3*...*k)*M$

We can say that:

$((k-1)*((k-2)*...(2+3) + 2*3) + 2*3*4)...) = C$

So that:

$k*C + (2*3*...*(k-1)) = (2*3*...*k)*M$

We can also say that:

$(2*3*...*(k-1)) = N$

So that our final equation is:

$kC + N = (2*3*...*k)M$

The series is known as the Harmonic series, and is often dealt with for primes.

If k is prime, this is the sum of a multiple and a non-multiple of k.

Clearly there is no M that satisfies the equation.

The Harmonic Series does not give an integer for any $k\geq2$

$Q.E.D.$

bjhopper · #18 $Old$ February 10th, 2008, 09:04 PM

two trains 100 miles apart are approaching at 30 miles per hour.they will collide after 3.33 hrs. the speed of the faster must be less than 30mph, the speed of the slower greater than zero neglect not moving.the bird wings it for 3.33 hrs so travels 200 mi . 60 x3.33hrs.
example train A 20 mph, train B 10 mph. they aapproach each other at 10 plus 20 mph .train Atravels 67 mi B 33mi