Is this right? With this formula I get 36 rectangles on a 4*4 grid while you get 30, but after drawing up the grid myself, I see that you can indeed put 36 and not 30 rectangles on a 4*4 grid, so my formula might actually work.
I don't get how you did that sum there, but a simpler way would be to simply pick 2 vertical lines and 2 horizontal lines:
2 vertical lines: m choose 2 = m(m-1)/2
2 horizontal lines: n choose 2 = n(n-1)/2
Multiply it out, and you have your answer...
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
fetofs wrote:
I don't get how you did that sum there, but a simpler way would be to simply pick 2 vertical lines and 2 horizontal lines:
2 vertical lines: m choose 2 = m(m-1)/2
2 horizontal lines: n choose 2 = n(n-1)/2
Multiply it out, and you have your answer...
Huh, yeah that is indeed a lot easier. :P I guess I made things a little too complicated.
Geometry puzzle:
1) Calculate the area of the shaded part.
2) Take the integral part of the result, negate it, and use it as longitude coordinates, in degrees. Take the two first digits after the decimal point of the area and use them as latitude coordinates in degrees. (Mathematically speaking, if the area is A, then latitude = floor((A - floor(A))*100) degrees.)
3) These coordinates will point to a state in the US. Y = the year when the capital city of this state was founded.
4) Take Y modulo 100 (iow. the last two digits) and substitute the 60.3 in the picture with this value.
5) The answer to the problem is the height h using this modified value, with at least 5 decimals of accuracy (after the decimal point).
A lot of numerical computation... I presume they used a computer to do this, too.
If you look at the right triangles on the circle, you can quickly see that (I would include a drawing if I could, but anyway)
h² = 8(2r-8)
h² + (2r-8)² = (60.3)²
This gives r = 32.2163 and h = 21.2474
The shaded area is equal to (r² * arctan(h/(r-8)) - h*(r-8)) / 2, which is 116.47...
47º N 116º W is inside Idaho, and Boise was founded in 1863. Therefore, substitute 63 for 60.3 in the equations above, and you'll get the value of h.
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
Here's something that might keep you entertained for a few minutes if you don't already know the answer (I thuoght about this when I woke up this morning):
Let n=2,3,4,... and let p be any prime >=2. Show that the n:th root of p is irrational for all these values of n and p.
It's not very hard to solve, but I'd be interested in seeing if you solve it in another way than I did.
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
Assuming that the nth root of P is a rational number then there is a rational root m such that m = a/b where a and b are positive integers and have no common divisors.
Therefore
p^(1/n) = a/b
p = (a/b)^n
p = (a^n)/(b^n)
a^n and b^n are integers as well, and have no common factors. Since p is of the form q/r where r > 1 (b ^ n > 1 for b > 1 for b = 1, that would mean the root of a prime is an integer, which is false because primes have no factors), p is not prime. Therefore by contradiction the nth root of p cannot be rational.
This is probably the run of the mill solution.
Joined: 10/28/2007
Posts: 1360
Location: The dark horror in the back of your mind
I don't claim any degree of rigour in the following, but here goes:
Suppose that the nth root of p is rational, and hence expressible in the form a/b for integers a and b. Let a_1, a_2, ... a_m be the prime factors of a and let b_1, b_2, ..., b_l be the prime factors of b.
Consider a^n; in the prime factorisation of this there will be n a_1s, n a_2s, etc.; similary in the prime factorisation of b^n there will be n b_1s, n b_2s, etc.
Now, a^n / b^n = p (by definition), or a^n = p b^n. This means that p is one of the prime factors of a^n. Per the above, there are n instances of it in a^n. For the equality to hold, there must be one less than a multiple of n instances of p in the factorisation of b^n. It has just been shown, however, that all of the prime factors in b^n appear n times. Where n >= 2, this is a contradiction.
Thus, the nth roots of a prime number for n >= 2 are irrational.
In many tabletop games where a d100 is required, it's a common trick to throw two (differently-colored) d10's instead, and use one of them multiplied by 10 and the other as is (because a physical d100 is very cumbersome and impractical to use). For example, if you throw a white d10 and a black d10, the white could be used as the tens, and the black one as the ones. The end result is the same as when throwing a d100: An even distribution. (Many die sets even contain a d10 with multiples of 10 instead of regular 0-9 numbers, for this exact purpose.)
However, suppose that you use a variant of this: Throw two d10's and then the larger result is always used as the tens, and the other as the ones. (So for example if you throw a 7 and a 2, then the result is 72, but if it had been a 2 and a 7, the result would still be 72.)
What is the probability distribution now?
I'm assuming d10 range from 0-9 and I'll also asume that 0 is smaller than 9. (even though rolling two zeros results in the largest number if i'm getting this right)
The average number that will be rolled is 65,35 then. 2% chance of rolling each of any two-digit numbers with the first one being higher, 0% chance of the opposite, 1% chance of rolling each of any numbers with equal digits.
hu? That's a simple map from the events of the first example (with known probabilities) to the events of the second example. Giving 1/100th for every repdigit (11, 22, ..), 2/100th for every ab with a>b and 0 for everything else. Any special insights you're looking for?
Joined: 1/16/2008
Posts: 358
Location: The Netherlands
when plotted with 'arbitrarily' chosen bins....
[URL=http://img18.imageshack.us/i/d10x101.jpg/][/URL]
if you plot it with bins of width 1 its a bit useless to look at...
edit: ps I assumed roll(0,0) = 0
Each probability of the following would be exactly even with every other probability (except the doubles which are halved:)
00 (or 100)
10, 11
20, 21, 22
30, 31, 32, 33
40, 41, 42, 43, 44
50, 51, 52, 53, 54, 55
60, 61, 62, 63, 64, 65, 66
70, 71, 72, 73, 74, 75, 76, 77
80, 81, 82, 83, 84, 85, 86, 87, 88
90, 91, 92, 93, 94, 95, 96, 97, 98, 99
You could *fit* a curve to this, but it's not really a curve. Pretty much you've hacked half of half of the possibilities and what you have left with a very discrete thing.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
As many of you probably know, the exponential function is larger than any polynomial for sufficiently large x values. To illustrate this, let's say we have an arbitrary polynomial of degree n, P(x) = a(0) + a(1)*x + a(2)*x^2 + ... + a(n)*x^n. Find an x0 such that exp(x)>P(x) for all x>=x0.
Let A=max(a(i)).
P(x) = Sum(a(i)*x^i) < Sum(A*x^i) = A*(x^(n+1)-1)/(x-1) < A*x^(N+1) if x > 2.
A*x^(n+1) < x^(n+2) if x > A.
Lastly, if x > exp(n+2), exp(x) > exp(exp(n+2)) > exp((n+2)^2) > exp(n+2)^(n+2) = x^(n+2), since exp(x) > x^2 is true for all x > 0.
So, we have exp(x) > x^(n+2) > A*x^(n+1) > P(x) if x>2, x>A, and x > exp(n+2)
Thus, exp(x) > P(x) if x > max(2, A, exp(n+2)) = max(a(i), exp(n+2)).
So, x0 = max(a(i), exp(n+2)) is a solution.
Mind you, it's not exactly the best solution, seeing as exp(x0) will be at least 200 times larger than P(x0) for any choice of P. I suspect this large discrepancy comes from x > exp(n) being a rather poor lower bound on exp(x) > x^n. Can anyone improve on this solution?
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
I did it this way, and I got a really bad (but different) approximation, but it still holds:
Throughout this whole argument, I will assume x>1.
exp(x) = Sum[(X^k)/k!, k=0,1,...)] > (pick out the n+1:th term from the taylor series) > (x^(n+1))/(n+1)!
For the polynomial we have:
P(x) = Sum(a(i)*x^i) < Sum(abs[a(i)]*x^i) < Sum(abs[a(i)])*x^n
Combining this we get
exp(x) >(x^(n+1))/(n+1)! > Sum(abs[a(i)])*x^n > P(x)
We want to find an x such that the middle inequality holds. Dividing it by x^n, we get that exp(x)>P(x) if
x > (n+1)!*Sum(abs[a(i)]) =: x0
I would be interested in seeing more approximations.
1) Calculate the area of the shaded part.
2) Take the integral part of the result, negate it, and use it as longitude coordinates, in degrees. Take the two first digits after the decimal point of the area and use them as latitude coordinates in degrees. (Mathematically speaking, if the area is A, then latitude = floor((A - floor(A))*100) degrees.)
3) These coordinates will point to a state in the US. Y = the year when the capital city of this state was founded.
4) Take Y modulo 100 (iow. the last two digits) and substitute the 60.3 in the picture with this value.
5) The answer to the problem is the height h using this modified value, with at least 5 decimals of accuracy (after the decimal point).
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if you plot it with bins of width 1 its a bit useless to look at...
edit: ps I assumed roll(0,0) = 0
