Math Is Fun Forum

Well, don't be too happy just yet, I'm not convinced that I'm right either .  I was just checking your algebra.  For one thing I think you need to prove your assumption that the greatest sum is achieved when xi alternates between 1 and 0, unless this is from some theorem that I'm unaware of.

First, you'll need to break this down into cases where n is even and when it's odd.  You could use the floor function, but I'm not comfortable with it and prefer just to do the work twice.  If you're ok with it then go ahead.

Your work assumes an even n, so let's go with that.  Your 3rd to last line is the same thing I got, so it appears to be correct.  Going from that to the next line you skipped several steps of work, and this is where the error is.  Try substituting n=6 in those two lines.  On the 3rd to last line you get 6 + 2[4 + 2 + 0] = 6 + 2[6] = 18.  Using n=6 on the 2nd to last line you get 6 + 2[ { (6 - 2) + 2} * { (6 - 2) / 2} ] = 6 + 2[6 * 2] = 30.

My only advice here is to be deliberate when you use something like summation, especially when you're dealing with non-trivial values.  Here is how I would follow your 3rd to last line:

You should be able to work it from here, just be careful when you separate out pieces of the summation.

If you want to compare final answers I get

Not exactly, the law of gravity doesn't break.  It's just that your formula, which follows Newtonian physics, is incorrect.  It's an accurate approximation when you apply it to systems the size of planet or galaxies, but at the atomic level it is no longer accurate.  That's when you have to start using things like quantum mechanics, and if you try to take r all the way down to 0 you get into uncharted territory (GUTS, string theory, and the like).

I'm not sure if this is easier or harder than what you're trying, but I immediately thought of integration when I read this problem.  What I'm thinking is something like this: Fix point A to be at the top of the circle, which is legal thanks to mathsyperson's logic.  Then imagine that we start point B to the top of the circle too.  We then move it incrementally around the circumference of the circle.  At every increment we calculate the probablility that point C, if randomly placed, will form an acute angle with A and B.  Add these all up and you have your answer.

In mathematical terms we want to integrate around the circumference of the circle, so 0 to 360 degrees (or 2pi, whichever you prefer).  Using mathsyperson's logic again we can simply integrate from 0 to 180 and multiply it by 2 since the rest is just a reflection of the first half.  All that's left is to determine what function we're integrating.

What you want to integrate is a function of the probability that C, if randomly chosen, will form an acute angle BAC.  To get an idea of what we're dealing with let's make a mental picture.  We have a circle with point A being fixed to the top, which is equivalent to saying that its position is 0 degrees/radians.  B is positioned at some point on the right half of the circle, with a position between 0 and 180 degrees.  Where on the circle can we place C such that BAC is acute?

It's easier to determine where BAC is obtuse, so let's do that and then subtract that probability from 1 to get the probability that BAC is acute.  First, you can clearly see that C must be on the left half of the circle in order for BAC to be obtuse.  You can also see that the degree of BAC will increase as C approaches A.  As long as B is not at exactly 0 or 180 degrees there is a point on the left half of the circle where BAC will be equal to 90 degrees.  If C is closer to A than this point then BAC is obtuse, otherwise it is acute.  I will now point out that BAC forms a right angle if B and C lie on the same diameter.  This should be easy enough to prove, and from this point on you can solve the problem.

If you want a solution to check your answer against I came up with 75%.  Maybe one of the smarter people here can confirm or deny that result.

Just a note, a riddle like this would probably be better suited for the "Puzzles and Games" board.

For this problem you want to use integration by parts.  Interestingly, I looked it up on Wikipedia to refresh my memory on how it is done and found that they use this exact problem as an example.  Here is the link: http://en.wikipedia.org/wiki/Integration_by_parts

For number 2 consider what is required for a number to end with a 0.  All numbers ending in a 0 are divisble by 10, which is 2 X 5.  100 is 2^2 X 5^2, etc.  That means that in order to determine the number of 0's at the end of 2008! you need to calculate the number of pairings of 2 and 5 you can make with all of the numbers from 1 to 2008.

Jane had a nice algorithm for just this sort of problem a couple weeks ago, if you search the forums you can find it.  If you'd like to figure it out on your own first you need to realize that there are more 2's in the prime factorization of 2008! than 5's, so we can focus just on finding the number of 5's.  In order to do that first determine the amount of numbers from 1 to 2008 that have 5 as at least one of their prime factors.  Then you figure out how many numbers in that list have 5 as at least 2 of their prime factors, and you add them together.  You then repeat this step with numbers that have 5 as at least 3 factors, then 4, and so on until you reach the point that there are no numbers with more factors of 5.

Edit: In case the explanation was unclear, which I think it was, here is an example.  How many 3's are in the prime factorization of 30!

First we figure out how many numbers have at least one 3 in their factorization.  In this case we have 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, which is ten.  We then find how many numbers from 1 to 30 have 3 as two of their prime factors: 9, 18, 27, so there are three of them.  Add this to the original ten to get 13.  Now we check for numbers with three 3's: 27.  That brings us to fourteen 3's in the prime factorization of 30!.  If we try checking for numbers with four 3's we find that there are none since 81 is the smallest such number, so we know that there are exactly fourteen 3's in 30!.  Now just do the same thing for 5's in 2008!.

I'd guess these were your steps:

The key here is moving from step 2 to step 3.  You're multiplying by 8a, which is positive when a is positive and negative when a is negative.  Whenever you multiply or divide both sides of an inequality by a negative number you need to switch the sign around.

To show you what would happen we'll split the problem into 2 sections: first is where a > 0, the second is where a < 0 (it should be obvious what happens when a = 0).  If we assume a > 0 then we leave the sign as it is and can use the proof I gave above.  Let's see what happens if we assume a < 0:

Thus we've found that a > 7/4.  However, we assumed that a < 0.  This is a contradiction, which means that there is no solution when a < 0.  You can see that a = 0 is not a solution either, so we're left with 0 < a < 7/4.

Perhaps one of the smarter people here will contradict me, but my guess is you just need to recognize by looking at it.  Something I've noticed when helping my friends with math problems is that I'm just able to "see" things, like factorizations or substitutions, that they don't.  Something like this problem just takes more practice (I certainly wouldn't have tried a factorization like that).  I would say it's just a matter of practice and learning to recognize when you can use tricks like the ones they used for this problem.

Edit: That being said, in this particular problem you can see that the main factorization is similar to the original expression.  I doubt you would be expected to be able to look at that problem and immediately see the correct factorization, but it would be reasonable for you to try things like (x + y + z)^2, (x - y - z)^2, etc. and finally work your way to the correct solution.

Ahhh, so that's why people keep asking.  I think that's the third time this week someone's asked that question.

First of all, no, I don't think there's an easier way of doing that problem.  It doesn't really add much work to the problem, you just have to go through the same basic procedure twice.

Secondly, your proof is flawed.  The proof of the odd n should look like this:

Not this:

I'm doing something wrong here, can anyone spot it?

Start by taking the log to remove the exponents.  Let's use base 2 for simplicity's sake.  For now we'll confine ourselves to values of x strictly greater than 0 (it's quickly proven that 0 is not a solution).

Now take the derivative of the left hand side to determine its critical points.

This would make x = 2 the only critical point, which can't be the case because both x = 2 and x = 4 are solutions.