I tried summing an expression involving binomial coefficients (in terms of the net number of cycles around the circle, the number of lefts, and the number of statisticians) over the number of lefts (considering only the case where there are at least as many rights as lefts, and treating the other case as symmetric), and I got an expression in Wolfram|Alpha involving a hypergeometric function, which it was able to express as an exponential when N=1, and when it could presume the remaining parameter was an integer.
Specifically, for a sequence in which the bill goes right around the circle k times, where k is a whole number, it takes m lefts and m+N*k rights, where m is a whole number; the probability of the bill taking this kind of sequence is nCr(2m+N*k,m)/3^(2m+N*k+1).
Wolfram|Alpha gave the sum over m from 0 to infinity (using "Binomial" instead of nCr) as F((N*k+1)/2,N*k/2+1;N*k+1;4/9)/3^(N*k+1), where F is the standard (2,1) hypergeometric function.
When I tried evaluating that directly, for the case N=1, I got ((3-sqrt(5))/2)^k/sqrt(5). Also, regardless of the value of N,
when k=0, both former and latter expressions are 1/sqrt(5).
There probably are some elegant identities with binomial coefficients that would allow you to calculate this directly, without involving a hypergeometric function first.
Anyway, it's easy to see that because N*k is also a whole number, that expression involving the hypergeometric function simplifies to ((3-sqrt(5))/2)^(N*k)/sqrt(5), and the probability is two times the sum of that expression over k from 0 to infinity, minus the k=0 case, to avoid double-counting.
It's a geometric sum, and it's then easy to see that the probability goes to 1/sqrt(5) as N approaches infinity; what looks trickier is that it seems like all of the probabilities are rational for finite values of N, but it looks like a slog to prove it.
For an infinite number of statisticians, one would have to compute , which I don't know how to do.
Wolfram Alpha gives the answer to this integral as 1/sqrt(5). I have no idea how it determined that.
I wouldn't know how to show directly that equals 1/sqrt(5); however, I can explain briefly where 1/sqrt(5) comes from.
Let p0 (you), p1 (person seated one away), p2 (person seated two away), ..., be the probabilities of winning depending on position. Now suppose that the number of people go to infinity. So pn=(1/3)pn+1 + (1/3)pn-1 for all n≥1. Supposing that pn+1=apn, plugging in to the equation above gives pn=(1/(3-a))pn-1. Without going into too much detail, applying f(x)=1/(3-x) to a number over and over will cause it to converge to a solution of x=1/(3-x); this gives x=(3-sqrt(5))/2 (since x has to be less than 1). So p1 approaches ((3-sqrt(5))/2)p0. Plugging this into p0=1/3 + (2/3)p1 gives p0=1/sqrt(5).
Edit:
arflech wrote:
It's a geometric sum, and it's then easy to see that the probability goes to 1/sqrt(5) as N approaches infinity; what looks trickier is that it seems like all of the probabilities are rational for finite values of N, but it looks like a slog to prove it.
Well, if N is finite, the probabilities are obviously all rational since the problem boils down to solving a linear system over the rational numbers. Which is a neat way to prove that the above quantity is rational for all N≥1.
It's a geometric sum, and it's then easy to see that the probability goes to 1/sqrt(5) as N approaches infinity; what looks trickier is that it seems like all of the probabilities are rational for finite values of N, but it looks like a slog to prove it.
Given the square roots of 5, I would wager that the proof would probably involve Fibonacci numbers and the golden ratio.
It sure does. We can write the above as:
For odd N, this equals FN/LN, while for even N, it is LN/5FN.
Anyway, it seems that the way to compute the integral is to make the substitution u = tan(ω/2), getting:
Now, what about the probability of the guy M spaces to your right winning? This should be obtainable by multiplying each term in the sum or integral by , or cos Mω, respectively.
Edit: Oh, I see, it was answered above for an infinite number of statisticians, and it can be written simply as .
Edit 2: Ok, it's obviously , so FN-2M/LN for odd N, and LN-2M/5FN for odd N.
I made this cute geometry problem for you 😛😈 Can you find the value of this angle (DAB)?
60 + 40 + phi + x = 180 (large triangle)
30 + 15 + y = 180 (bottom triangle)
30 + phi + z = 180 (left triangle)
25 + w + x = 180 (right triangle)
y + z + w = 360 (angles about D)
5 variables, 5 functions. But the solutions are insufficiently constrained with just these equations. And phi can take any angle in (0, 80) and still be valid.
phi elem (0,80)
z=150-phi
x=80-phi
w=phi+75
y = 135
So your question, unless I'm missing a hint, is ill-formed and has no unique solution.
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Amaraticando wrote:
I made this cute geometry problem for you 😛😈 Can you find the value of this angle (DAB)?
Let's start by labelling the angles as such: DAB = a, DAC = b, ADB = c, ADC = d, BDC = e
We have:
a + c + 30 = 180
b + d + 25 = 180
e + 30 + 15 = 180
a + b + 30 + 30 + 15 + 25 = 180
c + d + e = 360
These simplify down to
a + c = 150
b + d = 155
e = 135
a + b = 80
c + d = 225
So, c = 150 - a and b = 80 - a
d = 155 - b = 155 - (80 - a) = 75 + a
But then c + d = (150 - a ) + (75 + a) = 225, which we already knew.
So, no, you can't find that angle.
But something seems wrong about this. I think I'm missing an equation in those original ones...
EDIT:
Right, try 2...
So, let BC = 1.
Then, by the sin law:
BD = sin(15)/sin(135) = 0.366025404
AB = sin(40)/sin(80) = 0.652703645
Then, by the cos rule:
AD^2 = BD^2 + AB^2 - 2*BD*AB*cos(80)
AD = 0.382359928
Then, again by cos rule:
cos(DAB) = (AD^2 + AB^2 - BD^2)/(2*AD*AB)
cos(DAB) = 0.878011324
Therefore DAB = 28.59660668 degrees.
No, you guys forgot the Law of sines (or something similar).
Then (after cancelling out the lengths of the sides) you will end up with something like
sin(15°) / sin(25°) = sin(Φ) / sin(80°-Φ)
which looks easy... but isn't.
(oh whoops it is... look reply below by FractalFusion)
The solution is almost 180°/(2*pi), which I should write as (180°/pi)*0.5.
I say almost because the factor at the end is not actually 0.5. It's actually kind of a long constant, which roughly equals 0.499105.
So my best bet is ~28.597°
Though, I'm unsure myself. Maybe I just overcomplicated things.
Edit: <PJBoy> Masterjun, your constant can be simplified to arctan((cos(65) - cos(95)) / (2 sin(25) + sin(95) - sin(65)))
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
No, you guys forgot the Law of sines (or something similar).
Then (after cancelling out the lengths of the sides) you will end up with something like
sin(15°) / sin(25°) = sin(Φ) / sin(80°-Φ)
which looks easy... but isn't.
How to derive it quickly: sin(15°)/BD = sin(30°)/CD, sin(30°)/AD = sin(Φ)/BD, sin(80°-Φ)/CD = sin(25°)/AD. Multiplying all equations gives sin(15°)sin(30°)sin(80°-Φ)/(BD*AD*CD) = sin(30°)sin(Φ)sin(25°)/(CD*BD*AD), and cancelling and rearranging gives the above equation.
You can use the trigonometric form of Ceva's Theorem to get the above equation directly. Now that I think of it, the sine law gives a really fast proof of that theorem. Most other proofs I've seen use the sine area rule, which is a bit more involved.
By the way, we can solve the equation as follows:
sin(15°) / sin(25°) = sin(Φ) / sin(80°-Φ)
sin(25°)sin(Φ) = sin(15°)sin(80°-Φ) = sin(15°)(sin(80°)cos(Φ)-cos(80°)sin(Φ))
(sin(25°)+sin(15°)cos(80°))sin(Φ)=sin(15°)sin(80°)cos(Φ)
tan(Φ)=sin(15°)sin(80°)/(sin(25°)+sin(15°)cos(80°))
and Φ is obviously less than 90° so
Φ=arctan(sin(15°)sin(80°)/(sin(25°)+sin(15°)cos(80°)))
Correct! At a first glance, the problem looks trivial to solve. After not getting anything new, only a bunch of redundancies, it looks impossible, but the drawing makes it clear that the angle is determined by the given values.
There's a town where 70% of the men are married to 90% of the women. What fraction of the total town is married?
I had a suspicion that there was some background context related to this question, though upon searching the Web, I certainly didn't expect all the news articles regarding it. The fallout was amusing though.
This Riddler question has an obvious and straightforward extension, but the Extra Credit section uses a much more difficult extension: http://fivethirtyeight.com/features/how-big-a-table-can-the-carpenter-build/
I figured out approximately the maximal radius for N=3, and the N=4 case was easy, but I didn't find a real pattern, just a sense that the radius increases with N.
Here's a problem I thought about recently.
If you take the last digit of 102564 and put it first, you get 410256, which is 4 times the original number.
Similarly, if you take the last digit of 142857 and put it first, you get 714285, which is 5 times the original number.
Problem: What is the smallest positive integer such that if you take its last digit and put it first, you get a number which is 6 times the original? (As usual, numbers don't start with leading zeros.)
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Location: The Netherlands
FractalFusion wrote:
Here's a problem I thought about recently.
If you take the last digit of 102564 and put it first, you get 410256, which is 4 times the original number.
Similarly, if you take the last digit of 142857 and put it first, you get 714285, which is 5 times the original number.
Problem: What is the smallest positive integer such that if you take its last digit and put it first, you get a number which is 6 times the original? (As usual, numbers don't start with leading zeros.)
I got interested in this and wrote a quick program to brute-force all numbers to see if their digit-swapped versions were an integer multiple between 2 and 10. I didn't get an answer to the question asked, but I did find exactly seven results in the 100000-250000 range (one x5, the number FractalFusion noted, and the rest x4), and nothing else for at least the next five orders of magnitude (after this, the bot started taking too much time for my quick test).
Whatever the answer would be, if it even exists, it would be larger than at least the 10 billion range.
For completion's sake, the results I found:
102564 - 410256 (x4, FractalFusion's example)
128205 - 512820 (x4)
142857 - 714285 (x5, FractalFusion's example and the only x5 example)
153846 - 615384 (x4)
179487 - 717498 (x4)
205128 - 820512 (x4)
230769 - 923076 (x4)
http://www.youtube.com/Noxxa
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I found the following number:
1016949152542372881355932203389830508474576271186440677966*6 = 6101694915254237288135593220338983050847457627118644067796.
I still didn't check if it is the smallest. But it's surely the smallest that ends on a 6.
EDIT: this is indeed the smallest solution. The next smallest numbers end in 7, 8 and 9, and they are the product of this number and 7/6, 4/3 and 3/2, respectively.
Sorry for not trying to answer the question, but I got distracted and found these neat x2's
x2
473684210526315789
947368421052631578
or |||||||||||||||
368421052631578947
736842105263157894
or ||||||||||||
105263157894736842
210526315789473684
421052631578947368
842105263157894736
or ||||||||||||
263157894736842105
526315789473684210
or ||||||||||||||
157894736842105263
315789473684210526
631578947368421052
Also, it looks like there are an infinite amount of x5's, when you construct the numbers like:
142857
142857142857
142857142857142857
etc.
Oh, I just realized that this kind of concatenation also works on some of my x2's up there...
(uuuh, I guess it would make sense if it always worked?)
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
Here's a problem I thought about recently.
If you take the last digit of 102564 and put it first, you get 410256, which is 4 times the original number.
Similarly, if you take the last digit of 142857 and put it first, you get 714285, which is 5 times the original number.
Problem: What is the smallest positive integer such that if you take its last digit and put it first, you get a number which is 6 times the original? (As usual, numbers don't start with leading zeros.)
Spoilers contain explanation of working and notes on why it is so large (which makes no sense without the explanation).
Solving ax10n+b=6(10b+a) gives 59b=a(10n-6). Solving mod 59 (and removing the factor of a, a single-digit number and hence not divisible by 59) gives 10n-6=0, or 10n=6. Solving for n gives n=57.
Solving the resulting equation gives b=16949152542372881355932203389830508474576271186440677966 times a, and substituting a=6 gives the smallest value when it works (if a is smaller, b would not have enough digits):
1016949152542372881355932203389830508474576271186440677966 times 6 equals 6101694915254237288135593220338983050847457627118644067796.
The reason why it is so big is that 59 has no smaller divisors, whereas the similar numbers for 4 and 5 (39 and 49) have smaller factors that can be contained in a (3 and 7). the one for seven does (69(LOL) equals 3 times 23), but 8 and 9 are both prime, making even bigger numbers. 3 and 2 have smaller primes (19 and 29 respectively).
PS: When I started, only Mothayrays's post was in the forum. By the time I finished this, BrunoVisnadi posted the answer, so... nice to see we have the same answer.
FractalFusion and Masterjun, I am always happy to see people having fun with 142857, because it is easily one of my favourite numbers. (Well, more specifically the decimal expansion of 1/7.)
I am still the wizard that did it.
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Nice, you got the answer pretty quickly!
Yes, the smallest number is 1016949152542372881355932203389830508474576271186440677966. Bobthefloater's explanation works (though the way 'b' and 'a' are defined there means that b=169...796 instead of b=169...7966). I wrote it in a slightly different way.
Any such number N must satisfy 6*10N-N=A*(10n-1), or 59N=A*(10n-1), where A is the last digit of N and is a number from 1 to 9, and n is the number of digits of N. Therefore 59 divides 10n-1. The smallest n for which this works is n=58 (10 is a primitive root mod 59). So N=A*(1058-1)/59=A*169491525423728813559322033898305084745762711864406779661. Since N has to be 58 digits long, the smallest A that gives this is A=6, and that gives N=1016949152542372881355932203389830508474576271186440677966.
There is also a pencil-and-paper explanation as well, which I think BrunoVisnadi and Masterjun recognized. Let N be the number we want to find. The first digit of 6N has to be at least 6 (otherwise N has less digits than 6N) and N starts with 1. This means the second digit of 6N is 1. We have:
61.....
1......
The next digit of N to find is simply the remainder of the previous division by 6, joined to the corresponding digit of 6N (which is always the same as the previous digit of N). For example:
61.....
1......
(1 divide 6 is 0 with remainder 1)
610....
10.....
(10 divide 6 is 1 with remainder 4)
6101...
101....
(41 divide 6 is 6 with remainder 5)
61016..
1016...
and so on. The process terminates when a division has a remainder of 0 and the digit of N is 6 (the same as the first digit of 6N). This gives N=1016949152542372881355932203389830508474576271186440677966. Starting with 7, 8, or 9 as the first digit of 6N will give other numbers with the same number of digits, and they are larger. In fact, they are just cyclic shifts of the above number.
For x2 to x9 the smallest positive integer with this property is:
2: 105263157894736842
3: 1034482758620689655172413793
4: 102564
5: 142857
6: 1016949152542372881355932203389830508474576271186440677966
7: 1014492753623188405797
8: 1012658227848
9: 10112359550561797752808988764044943820224719
Other than for x5, the numbers resemble the repeating part of the decimal expansion for 2/19, 3/29, ..., 9/89.
I considered numbers which give strict multiples when their first digit is moved to the end, but that isn't as interesting. Only x3 works and the numbers are 142857, 285714, 142857142857, 285714285714, etc.
Invariel wrote:
FractalFusion and Masterjun, I am always happy to see people having fun with 142857, because it is easily one of my favourite numbers. (Well, more specifically the decimal expansion of 1/7.)
It's one of my favourite numbers too. I was familiar with such decimal expansions since childhood. (e.g. 1/7, 1/17, 1/19, ...)
, which I don't know how to do.
It's a geometric sum, and it's then easy to see that the probability goes to 1/sqrt(5) as N approaches infinity; what looks trickier is that it seems like all of the probabilities are rational for finite values of N, but it looks like a slog to prove it.
