Hot answers tagged

9

I can do it with only seven switches. I need to transmit three bits to my friend. Bit one is the parity of the number of switches up among $1,3,5,7$. Bit two is the parity of the number of switches up among $2,3,6,7$. Bit three is the parity of the number of switches up among $4,5,6,7$. I examine the initial setting, find the bits that are wrong, if any,...


6

The first player (P1) has a winning strategy, which I'll describe here. First, I'll quickly explain the notation I've used. Each 'cell' on the board has three lines passing through it, so I've categorised them by whether the cell is an end point or a midpoint of those lines: The cell is an endpoint of one line and a midpoint of the other two (those ...


4

I visualized the solution Nick provided: http://imgur.com/a/AuhnS (I found it because it had used this arrangement to build boats models with a magnet-game)


4

Choose four weights $a$, $b$, $c$, $d$, leaving $e$ out, and use three weighings to compare each pair of them against the other pair. There are three possible outcomes. One of the three weighings balances, say $a+d = b+c$, and in the remaining two weighings $a$ is always heavier than $d$. In the next weighing, pit $b$ against $c$. Without loss of ...


3

As mentioned by skyking, your definition does not work. Take any statement $P$ that is not provable (Your example of "$x \in S$" is one if you allow non-sentences). Then $( P \lor \neg P )$ is unconditionally true, but you would have assigned it the 'truth-value' $2$. So your definition is self contradictory. However, what you seem to want is Kleene's 3-...


3

The complete sparse Wichmann ruler $W(2,5)$ $$0, 1, 2, 5, 10, 15, 26, 37, 48, 59, 70, 76, 82, 88, 89, 90 $$ can generate all integer differences up to $90$. As $\alpha$ immediately also gives us $180^\circ-\alpha$, $180^\circ +\alpha$ and $360^°-\alpha$, this is a solution, though not necessarily minimal (it is minimal for a set giving all numbers up to $90$...


3

Notice that you can start from after the first step. Specifically, you only need to prove that, given an arbitrary matrix where each row is already in ascending order, sorting each column in ascending order keeps each row sorted. Here is a very neat trick to do this. Let $m$ be the number of columns. For each $k$ from $1$ to $m$, sort the rows of the ...


2

The four coins can be identified in at most $15$ one-coin versus one-coin weighings: Split the coins into groups $1, 2, 3$ with $4,4,3$ coins in the respective groups. Label the coins $(w_1,x_1,y_1,z_1),(w_2,x_2,y_2,z_2),,(x_3,y_3.z_3)$. Plan (barring early luck hits of equal weight) to investigate each of the groups in the following way: Do every pair ...


2

my hunch is this. Each weighing can turn out one of 3 ways -- be left side heavy, right side heavy or ballance. If we put the coins in a line, there are ${11\choose 7} = 28$ ways the counterfeits could be arranged. But each coin could be one of 8 different weights! I think there are $28\cdot8!$ different distributions to identify. $n > \log_3 28\...


2

If a random variable $Y$ is a priori uniformly distributed in $[0,100]$, and you learn later that its value is in fact $\leq x$, then conditioned on this additional information it is uniformly distributed in $[0,x]$. It follows that the conditional expectation of $Y$ then is ${x\over2}$.


2

First you are rolling until you hit a $5$ or $6$, which has probability $2/6$, so the average number of rolls is the inverse $6/2$. Then you are rolling until you hit the other one, which has probability $1/6$, so the average number of rolls is the inverse $6/1$. Hence, the total number of rolls on average is $9$. So in order for the game to be fair, you ...


2

Basic approach. Conceding that I haven't looked too closely at this, I would proceed as follows: There are five shapes: circle, cross, diamond, heart, and slashed ring (the "don't" symbol). There are four textures: graph paper, polka-dotted, slashed, solid. Individually, for each of these properties—shape and texture—find the pattern followed by the ...


2

The answer is $17$, as students number $8$ and $9$ are wrong. To see this, note that if student $i$ is wrong, then student $ki$ must be wrong for every $k \ge i$. As these will not be two consecutive numbers, this cannot be the case. This means students $2$ through $7$ must be right. Given $pq$, with $p$ and $q$ coprime, if student number $pq$ is wrong, ...


2

Not sure if it answers your question, but you can define them as follows: $\color\red{\neg{x}}$ as $\color\green{2-x}$ $\color\red{{x}\vee{y}}$ as $\color\green{({x}+{y})\bmod3}$ $\color\red{{x}\wedge{y}}$ as $\color\green{({x}\times{y})\bmod3}$ $\color\red{{x}\implies{y}}$ as $\color\green{\neg{x}\vee{y}}$ $\color\red{{x}\oplus{y}}$ as $\color\green{[\neg{...


1

Ten different sums of pairs are given, and there are only ten pairs, so you know all the sums, and this allows you to deduce the ranks of the summed amounts (in fact also the amounts, but you don't need those). Let $x_i$ be the $i$-th greatest amount. We have $x_1+x_2=1500$ and $x_1+x_3=1400$, and thus $x_2-x_3=100$. We also have $x_5+x_4=400$ and $x_5+x_3=...


1

This general type of problem is called a "lost in the forest problem," and precise solutions depend crucially on the geometry of the forest (or, in your case, the desert.) You can find more information by Googling that term. The usual versions of the problem deal with a bounded forest that you're trying to reach the edge of. If I understand your problem ...


1

Let the integers be $a_1, a_2,\dots,a_N$. Now, consider the sums $$\begin{align} S_1&=a_1\\ S_2&=a_1+a_2\\ S_3&=a_1+a_2+a_3\\ \vdots\\ S_k&=a_1+a_2+\dots+a_k\\ \vdots\\ S_N&=a_1+a_2+\dots a_N \end{align}$$ Now, if someone of $S_k$ is $0\pmod N$ then we are done. If none of then is $0\pmod N$, then all the $S_i$'s can be $1,2,\dots,N-...


1

There are 4 different ways to get from the corner to a point in the opposite small face, as shown in the picture below. Note that the way described by AN, going below the cube, is a shorter way to get to your R/S point in the question, and that point is actually worse than the corner. For an AxBxC cuboid, if we denote the starting corner as (0,0,0), the ...


1

The farthest spatial point from one vertex is at the other end of the space diagonal which has a distance of $\sqrt{3^2+4^2+5^2}=\sqrt{50}$. There's only shortest path along the surface of a cuboid among these two points. By drawing a net of the cuboid, you can see how to find it out. The shortest path is the minimum among the following: $$\sqrt{a^2+b^...


1

Start with 50 red marbles in the first bag and 50 blue marbles in the second bag. The probability of drawing a red marble in this case is $\frac{1}{2}$. If we move a red marble from the first bag to the second bag, the probability becomes \begin{equation*} \frac{1}{2}\left(1 + \frac{1}{51}\right) \end{equation*} So when we keep moving red marbles from the ...


1

$$x_1=50$$ $$x_{n+1}=n\left(\frac{x_n}{2}+5\right)$$ My python code to verify it.


1

I'm not too clear on what you mean by "transform the layout isomorphically", but if by an isomorphism you mean a bijection of nodes which preserves the rows of three then there isn't one. The ordinary tic-tac-toe board has eight rows of three, whereas this board has nine rows of three.


1

I'll explain the thought process more explicitly. We can model random choice of tying with the following algorithm that ties knots one at a time: Label the spaghetti strings $S_1, \dots, S_{50}$. Begin with one end of $S_1$. Tie it to a random end of the $99$ remaining. If it makes a knot (there is only one end that does), then start again with any of the ...


1

The definition of a loop is in fact in agreement with your second row of diagrams. For example, if string $A$ ans ends $1$ and $2$, string $B$ has ends $3$ and $4$, and so forth, if I pair $(1,8)(2,3),(4,5)(6,7)$ plus any other pariings among the remaining $92$ ends, I get a loop consisting of strings $A,B,C,D$.


1

The area of circle of radius $\;r\;$ is given by the formula $\;\pi r^2\;$ ,so if we know this value is, for example $\;S\;$, then $$\pi r^2=S\implies \pi=\frac S{r^2}$$ and you get the value of $\;\pi\;$ (up to certain accuracy, of course).


1

We can consider bit sums mod 2. Number all switches, starting at 0, ending 39, their bit values being $d[i]$ for index $i$. Take relative prime number "steps", for example 2,3,5 calculate initial arithmetic bit sum for these subsets $$b = [2,3,5]$$ $$a_i= \sum_{k=0}^{\lfloor39/b_i\rfloor} d[k \cdot b_i] \hspace{1cm} \text{ where } a_i,d[j] \in \mathbb{Z}_2 ...


1

I'll assume that the times at which the signals change are uniformly distributed and Calvin doesn't know anything about the remaining lights; his only information is from the display at the current light. Then we need to find the expected waiting times $T_{ij}$ if Calvin has to pass $i$ lights and has $j$ uses of the magic wand left, where $1\le i\le 3$ and ...


1

If $U \sim Unif(0,1),$ notice that $Y = U^2 \sim Beta(.5, 1),$ which concentrates probability near $0$. Because the density function of $Beta(.5, 1)$ is $f_Y(y) = 1/(2\sqrt{y})$ for $0 <y <1,$ the point at which the densities of $U$ and $Y$ cross is $1/4$. Also, $P(U > 1/4) = 3/4$ and $P(Y < 1/4) = 1/2.$ More generally, $P(U > a) = 1 - a$ ...


1

I guess my idea is similar to Christian Blatter above. I did a R simulation with the assumptions that buses arrive promptly every 10 and 15 minutes, respectively. However, the interval time among buses is fixed with uniformly distributed starting times. I simulate buses between [0,10000] minutes and the guy arriving at the bus stop at time t in [50,9950] and ...


1

More mathematically, it can be done as : The minute hand moves 360 degrees in 60 minutes. This means that the angle of the minute hand is given by 6t, where t is number of minutes past midnight. The hour hand moves 30 degrees in 60 minutes. This means that the angle of the hours hand is given by 0.5t. The hands start together at midnight. The first time ...



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