July 20th, 2014, 04:40 PM  #21 
Global Moderator Joined: Dec 2006 Posts: 20,476 Thanks: 2039 
The correct answer is 25, given a reasonable interpretation of the question! Now you have to (a) identify the errors already made, (b) identify the intended interpretation, and (c) show that the answer, based on that interpretation, is indeed 25. I'll let this one run for a bit to see whether anyone cracks it.

July 22nd, 2014, 03:12 AM  #22 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions  
July 28th, 2014, 03:11 PM  #23 
Global Moderator Joined: Dec 2006 Posts: 20,476 Thanks: 2039 
Here's a gentle hint: it's not difficult to get the right answer (25), but it's very easy to make a slip and hence get the wrong answer (100).

July 28th, 2014, 06:01 PM  #24  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra  Quote:
But the primary solutions are for $k = 0$ and $k = 1$. Thus, solutions are: $$\text{For $k=0$:} \qquad x=\frac{a \pm \sqrt{a^2  4 k^2 \pi^2}}{2} \\ \text{For $k=1$:} \qquad x=\frac{a \pm \sqrt{a^2  4 k^2 \pi^2}}{2} \\$$ When we sum the four solutions we get $2a = 100$. So I'm now missing four solutions.  
August 3rd, 2014, 07:54 PM  #25 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications 
You get a as the sum of the 2 solutions for each k. So you have to find a such that as k gets too large, the solutions become complex, and the real solutions sum to 100 (don't forget that we are taking the sine, so I assume that they are not looking for complex numbers in this case). With a=25, for k>3, the solutions are no longer real. So k=0,1,2,3. So 4a=100, a=25. With a=100, k can go up to 15 so things don't add up because you would have 15a=100. Similar for a=50, a=100/3, and too few k for a=20. 
August 4th, 2014, 07:13 AM  #26  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications 
Correction: Quote:
Also, to be more clear, since each solution of k gives a sum of a: Letting k=0 gives a=100, but valid k are 0,1,2...14,15; giving 16a=100. Contradiction. Letting k=0,1 gives a=100/2, but valid k are 0,1,2...7; giving 8a=100. Contradiction. Letting k=0,1,2 gives a=100/3, but valid k are 0,1,2...5; giving 6a=100. Contradiction. Letting k=0,1,2,3 gives a=100/4, and valid k are 0,1,2...3; giving 4a=100. Works., so 25 is a solution. Letting k=0,1,2,3,4 gives a=100/5, but valid k are 0,1,2,3; giving 4a=100. Contradiction. Etc.  

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