125

For the more visually inclined, arrange all positive integers in a 5-wide chart, as follows: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 ....


68

18 pieces: In 9 parts: In 8 parts: In 7 parts: We need to be able to split it into 9 parts of 56, so it can't hurt to make 9 pieces of 56 and then split those further. Since we need to do better than 19 pieces, we can have at most 18 pieces. This means that most of our 56s are split into exactly two parts (we can have an extra piece for every 56 we don't ...


52

Yes, it's possible. Start with the various 2x2 squares. Ignoring symmetries of rotation and color swapping, there are 1 combination of 1 color, 3 combinations of 2 colors, and 2 combinations of 3 colors. It is simple to brute-force these 6 combinations to show that each can be changed to a 2x2 square of 1 color. For example: $$ {12 \choose 31} \to {2 2 \...


50

There is another way to do this problem, that doesn't involve any sort of conditional branching at all. It is in fact possible to set a fixed weighing schedule beforehand and still determine which ball is lighter or heavier in just 3 weighings. I'll explain how below. The gist of problems like these is, how much information can you get from the procedure ...


50

Is there a different way to label two 20-sided dice with positive integers, so that their sum is still a random number between 2 and 40 with these same probabilities? Answer: Details: More details: How I solved it, and how you can check that the solutions are correct:


48

I am assuming that So, consider Now,


48

My most sincere apologies for this. Really.


48

I have a hunch that the answer is Explanation: Continuing this way, we see that


45

It works when you use a really wide line:


39

After 40 moves by each player, 150 minutes have elapsed on each player's clock. So the average time taken for a move is exactly $150/40=3.75$ minutes, i.e. 3 minutes and 45 seconds. Question 1 The answer is yes. Proof: by a continuous version of the pigeonhole principle, there must be a move by one player or other (WLOG, say Anand's $n$th move) that takes ...


38

This is because


38



34

Just for fun: Actually, there is solution, which formally satisfies all the rules. You just need to walk through a wall! Hard, but possible!


34

It is impossible. Quite the same problem is "Seven Bridges of Königsberg", it was solved (proven) by Euler. Suppose you have drawn such a line and follow it from one room to another. Since you must use each door you must have a look at each room out of 5. What are these rooms? There will be 3 (at least) rooms you always go through - if you enter it you ...


33

Looks like: Thanks to @Gamow's comment, this number's maximality can be proved by self-contradiction of the assumption that it is not maximal. Any more dominos would cover all 64 squares. Assumption to be disproved: All squares can be covered with dominos.    A. As the top left corner must be covered, start with a horizontal domino there. (...


32

The probability is $1/2$. We have a permutation that maps each box to the box whose key it contains. Once we open a box, we can open the box it maps to. So, we can open all the boxes exactly if there is no all-steel cycle. Label the boxes $1$ through $100$. We denote the permutation in cycle format like $(31)(542)(6)$. To make this canonical, write each ...


31

Split this into three groups of four, A1, A2, A3, A4; B1, B2...; C1, C2... Each step here corresponds to one weighing. Weigh A against B. If A > B, then weigh A1, B1, and B2 against B3, B4, and C1. If the weights are equal, then one of A2...4 is heavier; weigh A2 and A3. If they are equal, A4 is heavier. If one is heavier, then that ball is heaviest. If ...


30

I believe this works as a short proof.


28

Answer: Argument: ============ ============ ============


28

The dimensions are and the tiling looks like this: Working out the dimensions of the rectangle is quite easy. We know its total area is $4209$ (i.e., $2^2 + 5^2 + 7^2 + 9^2 + 16^2 + 25^2 + 28^2 + 33^2 + 36^2$). This factorizes as $3 \times 23 \times 61$, and in order to fit in a square with a side length of 36, the rectangle must be $3 \times 23$ units ...


28

Someone suggested I put this as an answer, so here goes: Argument:


28

Here is a simple strategy of how they could do it Proof


27

The answer is This is because Now, to calculate the position for 24 frogs, I broke it up into 2 parts: Therefore


27

First choose $2014$ and $2016$. Average = $2015$. Now take the $2015$s. Their average is $2015$. Now choose $2015$ and $2013$. Average = $2014$. Choose $2014$ and $2012$. Average = $2013$. Note that we can keep on continuing this approach and end up with a situation like $1$, $2$ and $4$ in the end. From here, choose $2$ and $4$. Average = $3$. ...


27

Using only combinations of either single or double button presses: Using 1 press Using 2 presses Using 3 presses Using 4 presses Using 5 presses Total For anyone curious as to my internal process of coming up with these values, I made a quick chart of each type of press combo and how many variations there were for each type, with x indicating it has ...


27

First of all, It will probably make a difference So let's concentrate on the other ones We can already stop here; so


25

Let's think of each piece not as curved but as an L shape, with two sections at a right angle. Adjacent pieces in the track join in a straight line. If we pair off the touching sections, we see the track is composed of two-unit segments connected at right angles. Since horizontal and vertical segments alternate, there are an equal number of each. Moreover, ...


25

Yes there is a solution with a very simple strategy: Start in (1,1). Always go the right most square that's unvisited I'll try to illustrate it. I checked it by hand on an 9x9 board and a very nice pattern emerges that makes it clear it works on any X by X board. \begin{array}{c|cccc} \ &1&2&3&4&5&6&7&8&9\\ \hline 1 &...


24

It turns out that you need four weights, measuring 1, 3, 9, and 27 pounds. The trick here is that you can put weights on both sides of the scale. If you have an object of 2 pounds on the left side, you can place the 3-lb. weight on the right side and the 1-lb. weight on the left side for the scale to balance. Similarly, if the object weighs 5 pounds, you ...


24

Claim: If David Copperfield distributes $n\ge3$ cards with numbers $1,2,\ldots,n$ over the three top hats, then only the following two types of distributions make the trick work: Put $1,2,3$ into different hats; put every $k\ge4$ into the same hat as its residue modulo $3$. Put $1$ into the first hat; put $2,\ldots,n-1$ into the second hat; put $...


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