# Tag Info

12

Before entering, the mathematicians agree on a choice of representatives for real sequences when two sequence are equivalent if they are equal past some index ; and a re-labeling of $\Bbb N$ into $M \times \Bbb N$ where $M$ is the set of mathematicians. Once a mathematician $m$ is in the room, he opens every box not labeled $(m,x)$ for $x \in \Bbb N$, and ...

9

In order for pirates to make consistent decisions, we must start with two assumptions. First, pirates are all completely logical, and that they are all completely logical is 'common knowledge' (see joriki's comment below). Second, we must take a stance on the bloodthirsty vs. friendly issue (see Micah's comment above), and the problem isn't as interesting ...

7

Look at the equation $\frac{1}{x}=a$. We use Newton's Method to approximate the solution. Let $f(x)=\frac{1}{x}-a$. The standard Newton iteration gives $$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}=x_n -\frac{\frac{1}{x_n}-a}{-\frac{1}{x_n^2}}.$$ This simplifies to $$x_{n+1}=x_n(2-ax_n).$$ Remark: Note that only subtraction and multiplication are used. If we ...

7

Yes, $n$ black squares is minimal. No matter how you tile your grid, there will always be at least one black square in each row and in each column because adding a new tile always places a black square in both of the rows and columns in which the tile was added. The The best you can do is have one black line along the diagonal. Here's one way to achieve ...

6

As $27234702932=2^2\cdot181\cdot37616993$ where the last two factors are primes If $x=2X,y=2Y$ $$X^2-Y^2=181\cdot37616993$$ If $X^2-Y^2=p\cdot q$ where $p,q$ are primes, the possible cases for $X+Y,X-Y$ are $\pm pq, \pm 1$ and $\pm p,\pm q$ For example, if $X+Y=1,X-Y=pq$ The number of positive factors of $p\cdot q$ is $(1+1)\cdot(1+1)=4$ So, there ...

5

It is not possible. Assume it were. Then just before the last pour, you would have two buckets with equal mixes and one with an unequal mix. This violates the fact that the total quantities of each of the three colors are equal. Also, pouring from a bucket that has an equal mix into one that has an unequal mix can never fix the receiving bucket. Even with ...

5

Let $A$ be any compact set with area 5. Fix a point $P$ in $A$, and define the function $f:[0,1]\times[0,1] \to \mathbb Z_{\ge0}$ by $(x,y) \mapsto$ the number of lattice points covered by $A$ when $P$ is translated to $(x,y)$ Then the integral of $f$ over $[0,1]\times[0,1]$ is 5. So unless $f = 5$ almost everywhere, $f(x,y)$ must be $\ge$ 6 for some ...

5

Assumptions: Pirates are rational, greedy, and bloodthirsty. If the head pirate has multiple possible proposals in which he maximizes his profit, he will choose between them randomly, with a uniform probability distribution. The first two items on this list are common knowledge among pirates. Notation: Let $G$ be the number of gold coins. A proposal in ...

4

If the expected value of this game is $a$, then at a die roll of $X$ you have the choice of either collecting $X$ or paying a dollar and restart, which gives you an expected value of $a-1$. To maximize the expected value, you should take $X$ if $X> a-1$ and start over if $X\le a-1$ (it does not really matter what we do when $X=a-1$). We obtain therefore ...

4

No. Hint: color $6 \times 6$ squares like a checkerboard. Here is an example of a $18 \times 14$ board colored as I suggested. Each square is $6 \times 6$ and the rectangles are $2 \times 6$. The extension to $66 \times 62$ should be clear. Note that there are more black squares than white squares, but that every $1 \times 12$ placing will cover $6$ of ...

3

Note: in this answer, I have taken "inaccurate" to mean inconsistent with whatever philosophy governs their behavior. Well, going through them, we see that, when asked "Who are you?": A will answer A, as he is an accurate truth teller. B, in order to be inconsistent with the fact that he is a truth teller, can answer with anyone other than B, as he is ...

3

HINT: Let $n$ be the number of aliens on the ship. As noted in the comments, you must assume that $n\ge 2$. Suppose that A is an alien on the ship; how many of the others can it know? The largest possible answer is $n-1$, and the smallest is $0$. How many possibilities is that? Is it actually possible for one of the aliens to know $n-1$ of the others ...

3

A fair die is equivalent to a random integer generator on the set $\{1, 2, 3, 4, 5, 6\}$. Of course, the order of the integers doesn't matter, so neither do the arrangement of the numbers on the die. But to answer your question about why this arrangement, according to http://en.wikipedia.org/wiki/Dice, this is presumably so the 1, 2, and 3 faces of the die ...

2

In the absence of any other information (which is rarely the case in Minesweeper), clicking a $1$ implies a $\frac{1}{8}$ chance that any of the adjacent $8$ squares is a mine. Now suppose you click an adjacent cell and it reveals a $1$. At this point, you have something like: $$\begin{array} & \star & \star & \star & \star \\ \star & 1 ... 2 The annulus with radii r_1<r_2 has area \pi(r_2^2-r_1^2). So it requires r_2^2-r_1^2 milliliters of paint. When r_2-r_1=1, we have that r_2^2-r_1^2=(r_2-r_1)(r_2+r_1)=r_2+r_1=2r_1+1. Now, if your nth black strip is an annulus with radii r_1(n)=r+2(n-1) and r_2(n)=r_1(n)+1, and the paint required is$$2r_1(n) +1= 2r + 1 + 4(n-1)$$So, ... 2 If the inner radius of a ring is r and the outer radius is r+1, the area of the ring is \pi(r+1)^2-\pi r^2=\pi (2r+1), so the ring takes 2r+1 milliliters of paint. The next ring is 2cm greater in radius (one black, one white), so takes 2(r+2)+1 milliliters. To paint x rings starting at r takes 2r+1+2(r+2)+1+2(r+4)+1+\ldots ... 2 Take the logarithm that maps multiplication/division into addition/subtraction:$$\frac{x}{y}=e^{\log{x/y}}=e^{\log x- \log y}.$$x,y >0. Also, see my answer for multiplying natural numbers here: Advocating base 12 number system 2 If you bid a, the probability you win the item is a/1000. The expected amount you win is (a/1000)\times (a/2) where a/2 is the mean of the value given it is less than a. Then you receive 3/2 multiplied by this as a prize. This is bounded by 3a/4 so on average you make a loss of \ge a/4. Of course, there is a possibility you win something, ... 2 There is no good answer without making some assumption about the probability distribution of the price. If the price were exactly 0 it would fit the 0-1000 range and you should not bid. If you want to assume a uniform distribution, the problem can be solved. Let my bid be y and the price of the chest be x. My gain is$$g(x,y)=\begin {cases} 0 ...

2

For the game as described, Bob wins if $0 \le N \lt a$ as Alice cannot move. Alice wins if $a \le N \lt a+b$, as she can leave Bob something in the range $0$ through $a-1$. The pattern recurs modulo $a+b$, because whoever is winning can make sure two moves remove exactly $a+b$ stones. So Bob's winning positions are $0 \le N \pmod {a+b} \lt a$ and Alices ...

2

Assuming that outer layer dissolves at same rate at all surfaces. Also assuming your question "Exactly how many?" means "exactly how many cubes are left?" Lucky for us $n^3$ is the volume of a cube of edge n, meaning the cube formed by gluing together $n^3$ cubes has each edge made up of n small cubes. That is a massive hint. The dissolved cubes would be ...

2

This is actually a well-known problem, see Mathworld, or Wikipedia. Roger Penrose made the first solution in 1958, with curved walls, and Tokarsky made a solution in 1995 with straight walls. Although Penrose's solution is not smooth, I believe if you smooth out the corners it will still answer the question posed.

1

Let $x_i$ be the number of times you push button $i$. You'll only ever push it $0$ times or $1$ time, so be can take $x_i$ to be modulo $2$. You would like light 1 to go from OFF to ON. When we examine the chart in the link in your comments, we can only affect light 1 by switching buttons 25 or 36. This yields an equation: $$x_{25}+x_{36}\equiv1$$ so that ...

1

The assumption of independence is very unreasonable, so if we make it we are not talking about real weather. But as long as we remember this is just a puzzle, we can go on. Edit: (The next sentence is no longer relevant, previously it said probability $2/3$, probably because of the language of odds.) In your calculation, there is no $2/3$ probability ...

1

Firstly, I do not agree with comments which say that the assumption of independence is unreasonable. Of course if the weather stations were perfect, they would never be wrong and so of course they would be linked together by always giving the same correct prediction. What is independent are the reasons for the stations being wrong 2/5th of the time. Suppose ...

1

In case of 1 pirate, there's no debate. If there are 2, then the n1 gets all the money, n2 gets none. Because if n2 proposes anything less, n1 will not agree, hence proposal won't get clear majority, hence n2 will die and n1 will get everything nonetheless. (1000,0) If there are 3 pirates, then n3 just has to give n2 1 coin to buy his loyalty. n1 goes ...

1

There are two things you need to notice. First, statements one three and four are mutually contradictory. So the false statement must be one of those three. So both statements 5 and 6 are true. If Richard took drugs then Bill both did and didn't take drugs. Therefore Richard did not and Bill did. So by statement 2 (which we know to be true) Sam also ...

1

For any given $a$ and $b$, the set of $N$ for which Alice wins is determined by some congruence conditions on $N$ - in other words, it's a periodic set (with some period depending on $a$ and $b$). For example, if $a=1$, then Alice can win if and only if $N$ is not a multiple of $b+1$. (Her strategy is to leave a multiple of $b+1$ stones for Bob every time he ...

1

Hint: There are $N$ aliens, and assume one of the alien knows $K$ aliens. $(N>K)$ Can each alien know distinct number of aliens? Assume $A_1$ knows ${A_2,A_3 \dots A_{K+1}}$, now that $A_2$ knows $K$ aliens, the aliens who he knows also knows him.(Though confusing, its the fact -evil laugh-)

1

The money hall problem is much more straightforward with 1000 doors. Now, you choose 1, the presenter opens 998 doors which don't contain a prize, leaving 1 remaining unopened door which you didn't choose. You can now choose to either open your door (which you picked at the start, so a 1/1000 chance its the correct one) or you can open the other door the ...

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