How Can We Make 100! Divisible by 12^{49}?

[Awsom Guy]

1 mile. because you are in the middle.

 

[lmoo]

You can do slightly better than that. Go in a straight line for sqrt(2) miles in, let's say, the northwest direction, then travel south to the west edge of the 1 mile radius circle surrounding your starting point, walk along the southern half of the circle, and then go a mile north, so that you finish sqrt(2) miles northeast of where you started.

I haven't been able to prove this is optimal yet, or tried to work out the expected distance version, so those are still things someone could try.

We can do better than that, even!

chingkui's solution gives a total worst-case path length of $$2 + 3\pi/2 \approx 6.712$$ miles.

StatusX's gives $$2 + \sqrt{2} + \pi \approx 6.556$$ miles.

But if we consider these as part of a family of solutions, where:

1. We start by going in some direction between north and west (including north itself, but not including directly west), until we reach the tangent line of our unit circle's northernmost point.
   
2. We then follow whichever other tangent line we're on, until we get back to our circle.
   
3. We follow the circle around to its easternmost point.
   
4. We go directly north for one mile.

Our total path length is a function of $$\theta$$, where $$\theta$$ is the angle, in radians, between our initial direction and north. So chingkui's solution is for $$\theta = 0$$, and StatusX's is for $$\theta = \pi/4$$.

In general, for $$0 \leq \theta < \pi/2$$, $$f(\theta) = sec\theta + tan\theta + 3\pi/2 - 2\theta + 1$$

The derivative simplifies nicely for finding the minimum, and we end up with $$2sin^{2}\theta + sin\theta - 1 = 0$$

Our good result here is $$\theta = \pi/6$$, which gives a total worst-case path length of $$1 + \sqrt{3} + 7\pi/6 \approx 6.397$$ miles.

We still have to think about how to actually prove this type of path is optimal, though, if it is, and how to do the expected value.

 

[lmoo]

Following up on my previous post...

If we assume this family of paths also contains the path with the lowest expectation value, we can at least come close to pinpointing it, though it may be messy. The edge of the forest is equally likely to be in any direction from our starting point, so the key is the rate at which we "clear" different directions during the different segments of our journey.

We can break the journey into 5 parts:

1. Our initial 1 mile along a radius of the unit circle, during which we clear zero radians, since we're still inside the circle during this time.
   
2. Our continuation straight out from the center for $$sec\theta - 1$$ miles, during which we clear $$2\theta$$ radians, where $$\theta$$ is our initial angle of travel, compared to north. At a given distance $$x$$ into this portion we'll have cleared $$2cos^{-1}(1/(1+x))$$ radians (we're not clearing them at a constant rate, of course).
   
3. Our path along a tangent line back to the circle, for $$tan\theta$$ miles, during which we clear zero radians, since we're traveling along the border of the area we've already cleared.
   
4. Our path around the circle itself, for $$3\pi/2 - 2\theta$$ miles, during which we, of course, clear $$3\pi/2 - 2\theta$$ radians. For this segment, our radians accumulated will exactly match our miles, and they'll both be accumulated at a constant rate.
   
5. And our 1 mile straight shot to the north, during which we clear $$\pi/2$$ radians, reaching our total of $$2\pi$$ radians. At any given point $$x$$ within this segment, we've cleared $$2tan^{-1}x$$ radians.

Next we apply the proper boundaries to the segments, we normalize, and we take the derivative with respect to $$x$$. Now we multiply by $$x$$, and we integrate.

Our expectation value $$<x>$$, as a function of $$\theta$$:

$$<x> = ln(sec\theta + tan\theta)/\pi + (1 - \theta/2\pi)(sec\theta + tan\theta) + \theta^{2}/\pi - 2\theta + ln2/2\pi + 15\pi/16$$

And if anyone can solve this:

$$0 = (sec\theta - tan\theta)/2\pi + (1 - \theta/2\pi)sec\theta(sec\theta + tan\theta) + 2\theta/\pi - 2$$

...then we'll have our initial angle for a minimum expectation value on this type of path. Just by plugging in numbers, I'm getting an angle slightly smaller than that for our worst-case path, which makes sense. That one was exactly $$\pi/6$$ radians, or 30 degrees. Here, it's about 25 degrees, or $$5\pi/36$$ radians, yielding $$<x> \approx 3.8477$$ miles. But my $$\theta$$ for the minimum expectation value doesn't quite match up to my $$\theta$$ for the zero of the derivative of the expectation value, so I have some rounding error, or some other problem here (I'm using Excel).

I'm not convinced this type of path yields the lowest expectation value, though, especially given leg 3, the tangent path, where we're accumulating no radians at all. Maybe altering this portion could slightly lower our result.

 

[chingkui]

let see if I can resurrect this thread by posting a question...

This is an actual phone interview question from an internet search engine company:

Imagine you are in a 100 stories building, which has a very special electrical problem that above a certain floor, all electrical outlets are not functioning properly (below that floor, all electrical outlets are functioning properly though). However, you don't know which floor is the first one that the problem occur. Your job is to locate which floor the problem start, and do it with minimal effort. You are given 2 light bulbs, and each of them will light up when you use them on the floor with no electrical problem. However, when you use anyone of them on the outlet with problem, the bulb will burn and you are going to lose one light bulb. One obvious way to locate the floor the problem start is to start with floor 1, test if the bulb burn, and go to the next floor and test until you find the floor the problem start. Doing this way, if the 1st floor is where the problem start, you are very lucky and only take one test. However, if the problem only occurs, say, starting from floor 80, then you have to test 80 times before you find the problem. It is even worse if the problem only occurs on the 100th floor. You can do better than that, and your job is to propose the optimal strategy: to use the 2 bulbs to find the problematic floor so that no matter which floor the problem start, the maximum number of tests would not be worse than any other strategy in the worst case. (Your proposed strategy should be in the form: which floor to test first? If the bulb burns, which floor next? If not, then which floor? ...)

More challenge: what if you have 3 bulbs? How about 4? And... if you have as many as u wish, what is the worst case optimal number of trials? How many bulb minimum would u be using?

 

[CRGreathouse]

Imagine you are in a 100 stories building, which has a very special electrical problem that above a certain floor, all electrical outlets are not functioning properly (below that floor, all electrical outlets are functioning properly though). However, you don't know which floor is the first one that the problem occur. Your job is to locate which floor the problem start, and do it with minimal effort. You are given 2 light bulbs, and each of them will light up when you use them on the floor with no electrical problem. However, when you use anyone of them on the outlet with problem, the bulb will burn and you are going to lose one light bulb. One obvious way to locate the floor the problem start is to start with floor 1, test if the bulb burn, and go to the next floor and test until you find the floor the problem start. Doing this way, if the 1st floor is where the problem start, you are very lucky and only take one test. However, if the problem only occurs, say, starting from floor 80, then you have to test 80 times before you find the problem. It is even worse if the problem only occurs on the 100th floor. You can do better than that, and your job is to propose the optimal strategy: to use the 2 bulbs to find the problematic floor so that no matter which floor the problem start, the maximum number of tests would not be worse than any other strategy in the worst case. (Your proposed strategy should be in the form: which floor to test first? If the bulb burns, which floor next? If not, then which floor? ...)

More challenge: what if you have 3 bulbs? How about 4? And... if you have as many as u wish, what is the worst case optimal number of trials? How many bulb minimum would u be using?

First, with just two bulbs, you must adopt the following strategy: Hit a particular list of floors until you fail, then go back to one above where you last succeeded and go up one at a time. (Otherwise you risk not knowing the exact floor.)

The 'obvious' strategy is to split the problem space in half (multiplicatively), so go 10 floors up, then 20, etc. This gives a worst case of the 99th or 100th floor, 18 tries needed.

But we can do better (in the worst case), since we're being too generous on the first try: if the bulb fails on the 10th floor, we'll need at most 10 tries. So try the 15th floor, the 29th floor, the 42nd floor, the 54th floor, the 65th floor. the 75th floor, the 84th floor, the 92nd floor, and the 99th floor. Better, but this can probably still be improved -- the last jump should be smaller. 14 works, 13 doesn't:
14 27 39 50 60 69 77 84 90 95 99
13 25 36 46 55 63 70 76 81 85 88 90 91
so the optimal for 2 bulbs is 14 tries, for 93 to 106 stories.

If you have ceil(lg floors) or more bulbs, a binary search should be optimal. For 2 < bulbs < ceil(lg floors) you can build the solution recursively, I think. Given a list of optimal heights for a given number of tries at B bulbs, make a list for B+1 bulbs using the same procedure recursively. Since the first is a second-degree polynomial (k tries -> k(k+1)/2 + 1 stories), I wouldn't be surprised if the others could be expressed as higher-degree polynomials; but this I haven't checked into.

 

[chingkui]

First, with just two bulbs, you must adopt the following strategy: Hit a particular list of floors until you fail, then go back to one above where you last succeeded and go up one at a time. (Otherwise you risk not knowing the exact floor.)

The 'obvious' strategy is to split the problem space in half (multiplicatively), so go 10 floors up, then 20, etc. This gives a worst case of the 99th or 100th floor, 18 tries needed.

But we can do better (in the worst case), since we're being too generous on the first try: if the bulb fails on the 10th floor, we'll need at most 10 tries. So try the 15th floor, the 29th floor, the 42nd floor, the 54th floor, the 65th floor. the 75th floor, the 84th floor, the 92nd floor, and the 99th floor. Better, but this can probably still be improved -- the last jump should be smaller. 14 works, 13 doesn't:
14 27 39 50 60 69 77 84 90 95 99
13 25 36 46 55 63 70 76 81 85 88 90 91
so the optimal for 2 bulbs is 14 tries, for 93 to 106 stories.

If you have ceil(lg floors) or more bulbs, a binary search should be optimal. For 2 < bulbs < ceil(lg floors) you can build the solution recursively, I think. Given a list of optimal heights for a given number of tries at B bulbs, make a list for B+1 bulbs using the same procedure recursively. Since the first is a second-degree polynomial (k tries -> k(k+1)/2 + 1 stories), I wouldn't be surprised if the others could be expressed as higher-degree polynomials; but this I haven't checked into.

Hi CR, you are answering it way too fast My thought process went pretty much like what you outlined, first the 'obvious' strategy, then improve to get the optimal solution. But it took me way more time :grumpy:

I don't remember if I have solved the problem with more bulbs, it was many years ago... all i can remember is that I probably have concluded the generalization is not that difficult.

And now it is your turn to ask a question to keep this alive.

 

[Gib Z]

If CRGreathouse doesn't want to post a problem, here's one:

Find the units digit in the decimal representation of $$\left[ \frac{10^{20000}}{10^{100} +3} \right]$$ where [x] is the floor function of x.

 

[uq_civediv]

i don't actually have a new question if this turns out correct so feel free to post a new one

good old taylor series of a/x gives $$\frac{a}{x} = \frac{a}{b} - \frac{a}{b^2}(x-b) + \frac{a}{b^3}(x-b)^2 - \frac{a}{b^4}(x-b)^3 + ... + (-1)^n \frac{a}{b^{(n+1)}}(x-b)^n + ...$$

where a = 10^20000; b = 10^100; (x-b) = 3

any n-th element in the sum is thus $$(-1)^n \frac{10^{20000}}{10^{100(n+1)}} 3^n = 10^{(19900-100n)} 3^n$$

we notice that if 19900-100*n >= 1 the element has no effect on the unit digit. 19900-100*n==0 at n=199, if i list the values of neighbouring elements

198: +10^100 * 3^198
199: -1*3^199
200: +10^-100 * 3^200 ~ 0.0000265
201: -10^-200 * 3^201

198 ends with 100 zeros, in case 199) 3^199=3^(49*4+3) == 7 (mod10) because 3^4==1(mod10), summing up to 199th (-3^199) leaves 3 as the last digit, and the remaining ones are again too small to affect the result that floor( ...sum... ) ends with 3.

 

[laohu]

If CRGreathouse doesn't want to post a problem, here's one:

Find the units digit in the decimal representation of $$\left[ \frac{10^{20000}}{10^{100} +3} \right]$$ where [x] is the floor function of x.

This one has already been answered, but here's an alternate solution:

$$\lfloor \frac a b \rfloor = \frac{a- a \mod b}{b}$$

We have that
$$10^{20000} \equiv (-3)^{200} = 9^{100} \pmod{10^{100}+3}$$

The units digit is then the last digit of
$$\frac{10^{20000}- 9^{100}}{10^{100}+3}$$
i.e.
$$(10^{100}+3)^{-1}(10^{20000}- 9^{100}) \equiv 3^{-1}9 \equiv 3 \pmod{10}$$

The units digit is 3.

 

[Gib Z]

Those solutions are both correct (and shorter than mine). Heres another:

Find the minimum value of |sin x + csc x + tan x + cot x + cos x + sec x|
for real numbers x.

 

[StatusX]

One can check (say, by graphing the function), that:

f(x) = sin(x) + cos(x) + csc(x) + sec(x) + tan(x) + cot(x)

is never zero. So finding the minima of |f(x)| can be done by looking at all the local minima and maxima and seeing which is closest to zero.

If you define u = sin(x)+cos(x), then with a little work you can show:

f(u) = (u² - u + 2)/(u-1)

Now f'(x) = f'(u) du/dx, so extrema of f(x) occur where either f'(u) or du/dx are zero. Since:

u = sin(x) + cos(x) = sqrt(2) sin(x + pi/4)

We see that du/dx = 0 when u = +/- sqrt(2). Plugging this in, we find:

f( u = +/- sqrt(2) ) = 2 +/- 3 sqrt(2)

Next consider f'(u), which one can check is:

f'(u) = (u² - 2u - 1)/(u-1)²

This is zero for u = 1 +/- sqrt(2). We should throw out 1+sqrt(2), since u can never get that big, but the other one gives:

f( u = 1 - sqrt(2) ) = 1 - 2 sqrt(2)

So the three candidates for min |f(x)| are:

3 sqrt(2)+2, 3 sqrt(2) - 2, and 2 sqrt(2) - 1

Clearly the first ones bigger than the others. The second is also greater than the third by sqrt(2) - 1. Thus the answer is:

min |sin(x) + cos(x) + csc(x) + sec(x) + tan(x) + cot(x)| = 2 sqrt(2) - 1

 

[StatusX]

Here's another. Show that:

$$\prod_{m=1}^{N-1} \sin \bigg( \frac{ m \pi}{N} \bigg) = \frac{N}{2^{N-1}}$$

 

[snipez90]

Lemma: If $P_0 P_1 \cdot\cdot\cdot P_{n-1}$ is a regular polygon inscribed in a circle of radius 1, then $P_0 P_1 \cdot P_0 P_2 \cdot\cdot\cdot P_0 P_{n-1} = n$.

Proof: WLOG assume that the vertices of the polygon are the geometric images of the n-th roots of unity, and $P_0 = 1$. Take the polynomial $f = z^n - 1 = (z-1)\cdot\cdot\cdot(z-\varepsilon ^{n-1}),$ where $\varepsilon = \cos{\frac{2\pi}{n}} + i\sin{\frac{2\pi}{n}}$. Then $n = f'(1) = (1-\varepsilon)\cdot\cdot\cdot(1-\varepsilon ^{n-1}).$ Taking the modulus of both sides gives the desired result.

Now
$$1-\varepsilon ^ k = 1 - \cos{\frac{2k\pi}{n} - i\sin{\frac{2k\pi}{n} = 2\sin^2{\frac{k\pi}{n}} - 2i\sin{\frac{k\pi}{n}\cos{\frac{k\pi}{n} = 2\sin{\frac{k\pi}{n}\left(\sin{\frac{k\pi}{n}-i\cos{\frac{k\pi}{n}\right).$$

Hence $|1-\varepsilon ^ k | = 2\sin{\frac{k\pi}{n}}$ for k = 0, 1, ..., n-1, and the desired trig identity follows from this and the lemma.

 

[snipez90]

Problem:

Determine (with proof) for which $\alpha > 0$ and for which $x$ is the function $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by
$$f_\alpha (x) =\begin{cases}\frac{1}{q^{\alpha}}&\text{if }x =\frac{p}{q}\neq 0, gcd(p,q) = 1, q > 0\\ 0 &\text{if }x = 0\,\, \mbox{or x is irrational}}\end{cases}$$
continuous. What about differentiability?

 

[laohu]

Problem:

Determine (with proof) for which $\alpha > 0$ and for which $x$ is the function $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by
$$f_\alpha (x) =\begin{cases}\frac{1}{q^{\alpha}}&\text{if }x =\frac{p}{q}\neq 0, gcd(p,q) = 1, q > 0\\ 0 &\text{if }x = 0\,\, \mbox{or x is irrational}}\end{cases}$$
continuous. What about differentiability?

To me, there doesn't seem to be any $$\alpha > 0$$ that makes the function continuous (let alone differentiable). Choose any epsilon-neighborhood around a rational $$x = \frac p q$$, $$[a,b] = [x-\epsilon,x+\epsilon]$$ . Since we can make (b-a)2^k arbitrarily large, we can make the difference between $$\lceil a 2^k \rceil$$ and $$\lfloor b 2^k \rfloor$$ arbitrarily large. Choose an odd integer m in the interval between them. $$y = \frac m {2^k}$$ is in the epsilon-neighborhood. That in turn means that we can make $$|f(x) - f(y)| = |\frac{1}{q^\alpha} - \frac{1}{2^{k\alpha}}| = \frac{2^{k\alpha}-q^{\alpha}}{q^{\alpha}2^{k\alpha}}$$ arbitrarily close to $$\frac{1}{q^\alpha}$$ for any epsilon-neighborhood, making f discontinuous at x (unless you allow $$\alpha = \infty$$, in which case f is the zero function and therefore both continuous and differentiable. $$C^\infty$$ in fact ;) )

 

[Office_Shredder]

Hence the question of "for which x" is it continuous!

 

[laohu]

Ah, I missed that one. Well, I don't have that much time right now, so I'll try a quick argument: assume f (for any given alpha) is continuous on domain S. We see that any irrational for which f is continuous must have an epsilon-neighborhood not containing any rationals and vice versa. Also, given any rational numbers in S, we can modify the argument above to show that we can find an epsilon such that the punctured epsilon-nbhd around x only can contain rationals with denominators larger than 2/epsilon. Therefore $$S \cap \mathbb Q$$ must consist of isolated points. Using the observation about irrationals above, we must similarly have $$S \cap \mathbb Q$$ has no limit points

Checking the other direction, we should have that f is continuous on S iff $$S \cap \mathbb Q$$ consists of isolated points of S and has no limit points.

 

[Little ant]

A Q&A game is simple: One person asks a relevant question (it can be research, calculation, a curiosity, something off-the-top-of-the-head, anything ... as long as it's a math question) and other people try to answer. The person who posts the first correct answer (as recognized by s/he who asked the question) gets to ask the next question, and so on.

Let me start this off with a simple number theory problem :

What is the least number than must be multiplied to 100! (that's a factorial) to make it divisible by $12^{49}$ ?

(throw in a brief -couple of lines or so- explanation with the answer)

the number is 54, because 54

 

[Little ant]

100¡ is (2^97).(3^46), so we need a number, (2^1).(3^3)=54, and then 100¡ will be divisible by 12^49=(2^98).(3^49)

 

[BenVitale]

If you fly around the world at constant latitude, how far do you travel?

 

[LeonhardEuler]

If you call the degrees of, for example, northern latitude, $$\beta$$, then I'm getting the distance to be
$$2\pi R\cos{\beta}$$
With R being the radius of the Earth plus the distance of the plane above the ground. This is assuming a perfectly spherical earth, of course.

I did it by setting the z-axis to be along the line from the center of the Earth to the North Pole. Then, you have
$$x^2 + y^2 + z^2 = R^2$$
If the radius of the circle you travel along is r, then this says
$$r^2 + z^2 = R^2 \rightarrow r = \sqrt{R^2 - z^2}$$
Now just express z in terms of $$\beta$$ as $$R\sin{\beta}$$, so the distance is

$$d = 2\pi r = 2\pi\sqrt{R^2-R^2\sin^2{\beta}} = 2\pi R\sqrt{1-\sin^2{\beta}} = 2\pi R\cos{\beta}$$

Correct?

 

[BenVitale]

Earth circumference = 25,000 miles
Say your constant latitude L = 85 degrees

you only travel 25,000 cos(85 deg) = 2,175 miles.

 

[LeonhardEuler]

Here's a question based on a neat formula I once found in a very complicated way, and then found it in a simple way years later. Find a formula for
$$\sum_{n=0}^{n=N}\sin{n}$$

Obviously, the objective is to find a formula that doesn't require adding a number of terms that depends on N. (no $$\sum$$ sign).

If that's too easy for you, try
$$\sum_{n=0}^{n=N}n^k\sin{n}$$
for integer k> 0.

 

[LeonhardEuler]

Problem:

Determine (with proof) for which $\alpha > 0$ and for which $x$ is the function $f:\mathbb{R} \rightarrow \mathbb{R}$ defined by
$$f_\alpha (x) =\begin{cases}\frac{1}{q^{\alpha}}&\text{if }x =\frac{p}{q}\neq 0, gcd(p,q) = 1, q > 0\\ 0 &\text{if }x = 0\,\, \mbox{or x is irrational}}\end{cases}$$
continuous. What about differentiability?

Building off what laohu wrote, we know that f is discontinuous for all rational x other than 0. Just take y to be a nonzero rational number a/b, and then
$$|f(x) - f(y)| = |\frac{1}{q^\alpha} - \frac{1}{b^{\alpha}}|$$
for rational nonzero x, and
$$|f(x) - f(y)| = |\frac{1}{b^{\alpha}}|$$
for irrational x. Since the irrational numbers are dense, every $\epsilon$ neighborhood will contain some irrational numbers and so |f(x)-f(y)| will always be at least $\frac{1}{b^\alpha}$ for some x in the neighborhood.

Take a look at the irrational numbers and 0. Let $x_0$ be irrational or 0. Choose $$\epsilon > 0$$. Let M be the smallest integer bigger than $\frac{1}{\epsilon^\frac{1}{\alpha}}$. Let A be the set of numbers of the form $\frac{p}{q}$ with $q\leq M; p \neq 0, gcd(p,q) = 1, q > 0$. This is a finite set that does not include $x_0$, so there must be some $$\delta$$ for which $$(x-\delta, x+\delta)\cap A =\emptyset$$ For all x in this neighborhood, we have
$$|f(x_0) - f(x)| = |f(x)| < \frac{1}{M^\alpha} < \epsilon$$
because all rational numbers in this neighborhood have denominators less than M in lowest terms by construction of the set. So f is continuous for all irrational and zero x, and for all $\alpha$ > 0.

 

[faizands]

dang... u guys are just too smart! wait i mean... you guys are like in college right? ohhhh! that's why... I am in 8th grade :) i understood absolutely nothing bout everything that u guys were talking about... HAHHAHAHAHAH XD I am good at math... really good but this is just wayyy too hard xD can someone ask me like some... easy questions? i have no idea wut all those signs are :( BTW NICE TO MEET YOU ALL :)!

 

[LeonhardEuler]

It's ok, faizands, all of this would have looked like gibberish to me in 8th grade, too. I actually graduated college already, and I'm in graduate school now. You have plenty of time to learn this stuff. Welcome to PF!

 

[Tedjn]

Here's a nice problem. This one is quite fun and something that does not require any higher mathematics.

2010 Putnam Problem B3

There are 2010 boxes labeled B₁, B₂, ..., B₂₀₁₀, and 2010n balls have been distributed among them, for some positive integer n. You may redistribute the balls by a sequence of moves, each of which consists of choosing an i and moving exactly i balls from box B_i into any other box. For which values of n is it possible to reach the distribution with exactly n balls in each box, regardless of the initial distribution of balls?

 

[LeonhardEuler]

Here's a nice problem. This one is quite fun and something that does not require any higher mathematics.

2010 Putnam Problem B3

There are 2010 boxes labeled B₁, B₂, ..., B₂₀₁₀, and 2010n balls have been distributed among them, for some positive integer n. You may redistribute the balls by a sequence of moves, each of which consists of choosing an i and moving exactly i balls from box B_i into any other box. For which values of n is it possible to reach the distribution with exactly n balls in each box, regardless of the initial distribution of balls?

If it is possible to fill each box B_i with at most i-1 balls, then no moves are possible. Since B₁ must have 0 balls, then under this assumption, the only solution is n=0. Such an initial distribution will always be possible if the number of balls is less than or equal to
$$\sum_{i=1}^{i=2010}(i-1) = \frac{2009*2008}{2} = 201736\approx 1003.5*2010$$
so there is only hope of a solution for n>1003 (other than n=0)

Suppose n>1003. At least 1 box must have a number of balls not less than it's label. It will be possible to move all of these balls to B₁ in the following way (assuming they are not already there): Box B_i contains k*i + m balls, $k\in 1,2,3...; m<i$. Move i balls to B₁. Now move i-m balls to B_i from B₁. Then move the remaining k*i balls to B₁ in k steps. We have removed at least i balls from this box and placed them in B₁ (if they weren't already there).

If we remove this empty box (if it exists) from consideration, we can ask the question of whether it is possible that the remaining boxes can each have at most i-1 balls in each B_i, no matter how the balls from box 1 are distributed. This was not possible before the empty box was removed by construction. The removal of the empty box removes a capacity of (i-1) balls, while producing a surplus of at least i balls in B₁. Therefore, there is a way of distributing the balls from B₁ so that at least one box other than the emptied one has at least i balls.

Since balls in B₁ can be redistributed individually, it is possible to distribute them among the other boxes any way desired. Redistribute them as described above to give at least one B_i with at least i balls. Apply the previously described procedure of moving all the balls in this box into B₁. By an identical argument to the preceding paragraph, this procedure can be applied indefinitely until only box 1 remains. Move the balls individually so that each box contains n balls.

So the final answer is n=0 or n>1003.

edit:The question said positive integer n, so n=0 is not a solution

 

[chwala]

guys ,
what is the next question i think the question on c has not been answered,

ken

 

[waters68]

There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97.

 

[snipez90]

If a group G is the set theoretic union of a family of proper normal subgroups each two of which have only the identity in common, then G is abelian.

 

[silkplants]

very hard question but its interesting to solve, i will try it.

 

[MonkeyPwn]

There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97.

similiarly there are

33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess

2*3^50 will do

 

[snipez90]

Let X be normal, let a_1, a_2, ..., be in X, and let A = {a_1, a_2, ...}. Show that the following are equivalent:

I. No a_i is a limit point of A.
II. There is a continuous function f: X -> [0,1] such that f(a_i) = 1/(2^i) for all i.

 

[snipez90]

Let f be holomorphic on the unit disc.

a) If f(x,y) = u(x) + iv(y), what can you say about f?
b) If $f(x,y) = \phi_x + i\phi_y$ for a real function $\phi$, what can you say about f?