While -W(-1) is one solution (I'm not going to talk more about this; go see blackpenredpen if you're interested), this reasoning doesn't explain why there aren't any other solutions, since any z such that e^(e^z) = z (not necessarily e^z=z) could be a candidate solution for e^z = ln(z).
k | intersection
----------------------
-5 | 3.28777 +26.5805 i
-4 | 3.02024 +20.2725 i
-3 | 2.65319 +13.9492 i
-2 | 2.06228 +7.58863 i
-1 | 0.318132 +1.33724 i
0 | 0.318132 -1.33724 i
1 | 2.06228 -7.58863 i
2 | 2.65319 -13.9492 i
3 | 3.02024 -20.2725 i
4 | 3.28777 -26.5805 i
5 | 3.49852 -32.8807 i
We only need to consider when x>0, so the curves we need to intersect are:
y=x3
x2+(y-r)2=r2
The intersection point between the two functions is given as a polynomial equation: x6-2rx3+x2=0
Furthermore, there is a multiple zero at that point. So the x-derivative must also be 0: 6x5-6rx2+2x=0
After factoring x2 and 2x, respectively, and removing them (since x>0), we must solve the system of equations:
x4-2rx+1=0
3x4-3rx+1=0
Multiplying the first equation by 3 and subtracting the two equations results in the equation 3rx=2. Plugging that into the second equation and solving gives x=3-1/4 (about 0.759836), and r=2*3-3/4 (about 0.877383). This gives an area of the circle as pi*4*sqrt(3)/9 (about 2.418399).
I thought it was pretty interesting, and thought there would be other examples of matrices satisfying a similar property. It's barely possible to simply brute force all pairs of 2x2 matrices and search for the ones whose product is just digit concatenation, but out of boredom, I actually managed to come up with an algorithm to optimize this search using number theoretic concepts.
Anyway, now I have a list of all 2x2 matrices whose product is obtained by just concatenating their base 10 digits.
One thing that's very cool is that in some pairs the matrices are actually the same. Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.Anyway, now I have a list of all 2x2 matrices whose product is obtained by just concatenating their base 10 digits.
come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
You test all 104=10,000 possibilities to find the 153 4-tuples.
Then you can see that the same restrictions apply again for the lower right number, so exactly these 4-tuples appear again with flipped positions.
At this point you can just test all 153*153=23,409 candidates for the remaining 2 restrictions.
I found 101 solutions.
Then I figured I'd test the same approach with 2-digit number multiplication resulting in 4-digit numbers (with leading 0s). After searching the 1004=100,000,000 possibilities, I found 24,209 4-tuples, which means another 586,075,681 tests for solutions. After a minute of running it, I found 36494 solutions. One example is {{10,19},{23,44}}*{{14,38},{46,82}}:
In the 2-digit to 4-digit case, this reduces the tests needed in the second loop from 586,075,681 to 11,188,789, and takes about 10 seconds.Anyway, the problem is: come up with an algorithm, better than plain brute force, to find two single digit 2x2 matrices whose product is obtained by concatenating the base 10 digits of each element.
We want to find A = [a b / c d] and B = [e f / g h] such that AB = [10a+e 10b+f / 10c+g 10d+h], with a,b,c,d from 1 to 9 and e,f,g,h from 0 to 9. Note that AB-10A-B = [0 0 / 0 0], and so by adding 10I to each side (where I is the identity matrix [1 0 / 0 1]) and factoring, we have:
(A-I)(B-10I)=10I
Letting A'=A-I and B'=B-10I we have A'B'=10I. It follows that det(A') must divide 100. Furthermore, let A'= [a' b / c d'], where a'=a-1 and d'=d-1. Then B' is given by:
B' = A'-1 10I = (10/det(A')) [d' -b / -c a']
Note that the signs within B' must be [- + / + -] by restriction (it is impossible for them to be 0 and satisfy the restrictions on e,f,g,h), and so det(A')=a'd'-bc must be negative. Let k = -det(A'). Then we can solve for everything as follows:
a=a'+1 , b=b , c=c , d=d'+1 , e = 10 - 10d'/k , f = 10b/k , g = 10c/k , h = 10 - 10a'/k
Note that since det(A')det(B')=det(10I)=100, we have that k=bc-a'd' must be a positive divisor of 100. This restricts the possibilities greatly and now it is possible to enumerate everything by hand.
It turns out that k can only be 5, 10 or 20, and bc must be greater than k. Furthermore, if k=5, all of b,c,a',d' are between 1 and 4. If k=20, all of b,c,a',d' must be even. Finally, none of b,c,a'd' can be 0, and a' and d' cannot be 9. Because b and c are interchangeable, as well as a' and d', we classify the following according to b>=c and a'>=d', with up to four possibilities for each:
k bc b c a' d'
5 ==========
6 3 2 1 1 [2 3 / 2 2][8 6 / 4 8] [2 2 / 3 2][8 4 / 6 8]
8 4 2 3 1 [4 4 / 2 2][8 8 / 4 4] [4 2 / 4 2][8 4 / 8 4] [2 4 / 2 4][4 8 / 4 8] [2 2 / 4 4][4 4 / 8 8]
9 3 3 4 1 [5 3 / 3 2][8 6 / 6 2] [2 3 / 3 5][2 6 / 6 8]
3 3 2 2 [3 3 / 3 3][6 6 / 6 6]
10 ==========
12 6 2 2 1 [3 6 / 2 2][9 6 / 2 8] [3 2 / 6 2][9 2 / 6 8] [2 6 / 2 3][8 6 / 2 9] [2 2 / 6 3][8 2 / 6 9]
4 3 2 1 [3 4 / 3 2][9 4 / 3 8] [3 3 / 4 2][9 3 / 4 8] [2 4 / 3 3][8 4 / 3 9] [2 3 / 4 3][8 3 / 4 9]
14 7 2 4 1 [5 7 / 2 2][9 7 / 2 6] [5 2 / 7 2][9 2 / 7 6] [2 7 / 2 5][6 7 / 2 9] [2 2 / 7 5][6 2 / 7 9]
7 2 2 2 [3 7 / 2 3][8 7 / 2 8] [3 2 / 7 3][8 2 / 7 8]
15 5 3 5 1 [6 5 / 3 2][9 5 / 3 5] [6 3 / 5 2][9 3 / 5 5] [2 5 / 3 6][5 5 / 3 9] [2 3 / 5 6][5 3 / 5 9]
16 8 2 6 1 [7 8 / 2 2][9 8 / 2 4] [7 2 / 8 2][9 2 / 8 4] [2 8 / 2 7][4 8 / 2 9] [2 2 / 8 7][4 2 / 8 9]
8 2 3 2 [4 8 / 2 3][8 8 / 2 7] [4 2 / 8 3][8 2 / 8 7] [3 8 / 2 4][7 8 / 2 8] [3 2 / 8 4][7 2 / 8 8]
4 4 6 1 [7 4 / 4 2][9 4 / 4 4] [2 4 / 4 7][4 4 / 4 9]
4 4 3 2 [4 4 / 4 3][8 4 / 4 7] [3 4 / 4 4][7 4 / 4 8]
18 9 2 8 1 [9 9 / 2 2][9 9 / 2 2] [9 2 / 9 2][9 2 / 9 2] [2 9 / 2 9][2 9 / 2 9] [2 2 / 9 9][2 2 / 9 9]
9 2 4 2 [5 9 / 2 3][8 9 / 2 6] [5 2 / 9 3][8 2 / 9 6] [3 9 / 2 5][6 9 / 2 8] [3 2 / 9 5][6 2 / 9 8]
6 3 8 1 [9 6 / 3 2][9 6 / 3 2] [9 3 / 6 2][9 3 / 6 2] [2 6 / 3 9][2 6 / 3 9] [2 3 / 6 9][2 3 / 6 9]
6 3 4 2 [5 6 / 3 3][8 6 / 3 6] [5 3 / 6 3][8 3 / 6 6] [3 6 / 3 5][6 6 / 3 8] [3 3 / 6 5][6 3 / 6 8]
20 5 4 5 2 [6 5 / 4 3][8 5 / 4 5] [6 4 / 5 3][8 4 / 5 5] [3 5 / 4 6][5 5 / 4 8] [3 4 / 5 6][5 4 / 5 8]
24 8 3 7 2 [8 8 / 3 3][8 8 / 3 3] [8 3 / 8 3][8 3 / 8 3] [3 8 / 3 8][3 8 / 3 8] [3 3 / 8 8][3 3 / 8 8]
6 4 7 2 [8 6 / 4 3][8 6 / 4 3] [8 4 / 6 3][8 4 / 6 3] [3 6 / 4 8][3 6 / 4 8] [3 4 / 6 8][3 4 / 6 8]
25 5 5 5 3 [6 5 / 5 4][7 5 / 5 5] [4 5 / 5 6][5 5 / 5 7]
28 7 4 6 3 [7 7 / 4 4][7 7 / 4 4] [7 4 / 7 4][7 4 / 7 4] [4 7 / 4 7][4 7 / 4 7] [4 4 / 7 7][4 4 / 7 7]
30 6 5 5 4 [6 6 / 5 5][6 6 / 5 5] [6 5 / 6 5][6 5 / 6 5] [5 6 / 5 6][5 6 / 5 6] [5 5 / 6 6][5 5 / 6 6]
35 7 5 5 5 [6 7 / 5 6][5 7 / 5 5] [6 5 / 7 6][5 5 / 7 5]
40 8 5 6 5 [7 8 / 5 6][5 8 / 5 4] [7 5 / 8 6][5 5 / 8 4] [6 8 / 5 7][4 8 / 5 5] [6 5 / 8 7][4 5 / 8 5]
42 7 6 8 4 [9 7 / 6 5][6 7 / 6 2] [9 6 / 7 5][6 6 / 7 2] [5 7 / 6 9][2 7 / 6 6] [5 6 / 7 9][2 6 / 7 6]
45 9 5 7 5 [8 9 / 5 6][5 9 / 5 3] [8 5 / 9 6][5 5 / 9 3] [6 9 / 5 8][3 9 / 5 5] [6 5 / 9 8][3 5 / 9 5]
20 ==========
24 6 4 2 2 [3 6 / 4 3][9 3 / 2 9] [3 4 / 6 3][9 2 / 3 9]
32 8 4 6 2 [7 8 / 4 3][9 4 / 2 7] [7 4 / 8 3][9 2 / 4 7] [3 8 / 4 7][7 4 / 2 9] [3 4 / 8 7][7 2 / 4 9]
36 6 6 4 4 [5 6 / 6 5][8 3 / 3 8]
6 6 8 2 [9 6 / 6 3][9 3 / 3 6] [9 6 / 6 3][9 3 / 3 6]
In Magic: The Gathering the keyword "scry n" means "look at the top n cards of your library (ie. deck). You may put any amount of them on the top of your library in any order, and the rest on the bottom of your library in any order." For example, "scry 2" means that you look at the top 2 cards of your deck, and then you may choose to put both of them on the top of the deck (in either order), or both of them on the bottom of the deck (in either order), or one of them on the top and the other on the bottom. If you can get arbitrarily many consecutive "scry 2" effects (there are ways to achieve this), it can be relatively easily proven that you can use them to reorder your entire deck in any order you want. So my question is: Suppose your deck has currently 50 cards, and you can apply as many "scry 2" operations to it that you want. How many such "scry 2" operations do you need, at most, to reorder your entire deck in a particular order?
Prove that: 13 + 23 + 33 + 43 + ... + n3 = (1 + 2 + 3 + 4 + ... + n)2
(courtesy of artofproblemsolving.com)This is known already and there are probably hundreds of proofs out there (induction, difference of powers formulas, etc.)
I would like to repeat the question I already presented some years ago to a somewhat similar response: Do all the math challenges posted here need to be something for which nobody has ever found nor published a proof? Is this thread titled "some math challenges", or is it titled "unsolved problems in math"? And on that note, googling for the answer isn't much of a "math challenge". The only slightly challenging thing might be to find the correct search terms, but I don't think that's the type of "challenge" this thread is for.
I'm thinking your hypothesis is false, but I just have a thought, and that's far from a proof. A rational number can be written as x/y. By allowing negative a1 etc., you get the 1/y part. But what about a number x/y where you need the same number p1, p2, or p3 to make both x and y?
