Notice that each row and each column contains the numbers 1, 1, 2, 3, 4, 5, 6, 7  \begin{array}{ a b c d } 4*7 & 1*6 & 3*1 & 2*5 & \\ 3*5 & 2*1 & 4*6 & 1*7 &\\ ...

Each of the numbers 0,1,2…,9,10 appears at least once. In fact, the only way this works is if there are 1 students with name A, 2 students with name B, 3 students with name C, etc up to 11 students ...

Thanks to Alconja for the suggestion. My first thought upon seeing this problem is, I wonder what the RGB and HSV values of those bands are. The program I used gives R, G, B in the range 0-255 r g ...

Here is what I did: I started with these 3 sequences which are able to reproduce the number 1-216. Basically this is the first three digits of a base 6 representation. Each row represents one die. 1,...

When drawn on a hexagonal grid, it looks symmetric (2,8) red (2,1) (2,3) (7,8) (9,8) green (2,4) (2,6) (4,8) (6,8) (7,1) (7,3) (9,1) (9,3) blue (1,8) (2,7) (2,9) (3,8) (4,1) (4,3) (6,1) (6,3) (7,4)...

Partial solution. I can reduce this to a polynomial equation with only integral powers, but so far I am not sure how to find the integer solutions of this new equation. New equation: where m < 1,...

Are you: I'm not sure how this relates to the word culture though.

How about: I have teeth, but can be chewed. I alone am worshipped, but those who yield me are too. I reside among legends, but a myth I am not.

Your strategy should be to Edit: My comments The first result seems to imply that the end result of the competition will be Whereas the second result seems to imply that the very end will be: ...

Edit: I kept going back and forth between whether it should be 15 or 16 squares. I used color to convince myself that 15 does actually work. 1 is in the bottom left corner of a 8x8 square. All of the ...

I tried tackling the problem numerically. I found that up to 13 layers of depth(the furthest I could go), there was no solution. First I inverted $f(x) = 2x+1 -> f^{-1}(x) = \frac{x-1}{2}$ $... View answer 4 votes Obviously player two wants it to be so that after an even number of coins have been played no more can be. To do that, player two should prevent the forcing of an odd coin game and look for a way to ... View answer 4 votes The probability of Alice getting$i$heads after$n$flips is: The probability of Bob getting$i$heads after$2n$flips is: Then the probability of Alice and Bob getting the same number of heads ... View answer 4 votes I wrote a program to find general solutions to the decanting problem. I generalized the problem a little bit by allowing you to search for not just getting a single jug with$n$liters, but listing ... View answer 4 votes You can solve for n in at most Here is how this works: Furthermore: example: Edit, I realized this does not work for the case n =$3^4 * 5 * 11^4 * 13^4\$ because

I was able to do it in 6. 1. H H - - H T T T T T H H 2. H H T T H T - - T T H H 3. H - - T H T H T T T H H 4. H T H T H T H T T - - H 5. H T H T H T H - - T T H 6. H T H T H T H T H T

I would like to thank dmg and user3294068 because their insights helped me find this answer. dmg tried moving out twice as far in each step, whereas user3294068 tried moving w times as far in each ...

Edit: Angel Koh came up with another solution to my layout, so it is non-unique. I believe to have a unique solution in regular sudoku you need at minimum: 1 number in each column, 1 number in each ...

The number of moves it takes in order for the shortest alien to be at the front of the line is O(n). Therefore, they keep dancing until enough aliens die off that the dance converges in a reasonable ...

The answer is: To see why consider:

It must be because:

My best guess:

I believe the fathers are: Some of us with many Some of us with one Our children vary in appearance even more than we But helpers they are to one and plenty A few of us knew our calling A few ...

I believe you are: I seem physical.. On the fringe between two countries, I have no proper home, but make one for others. Standing proudly for all to see, my purpose is often to conceal. ...