New answers tagged

1

I am not giving a new solution. But I'd like to propose a nicer expression of the solution given for the general case $n=2^k$. In fact my solution is the same as noedne's. And the explanation why it works is the same as OHO's.


1

There is no need to guess, also proves that solution is unique. L stands for left sudoku, R for the right one. (X,Y) means the X is in L, Y in R, the rest should be self-evident. Formatting of sudokus is crappy but I believe it should be clear enough. Then L/R, horizontal lines are not drawn. Then Solution of L/R at this point is Now notice that At ...


0

CG has a correct and accepted answer, as verified by a python program I wrote that generates all possible chains over the range 001...360. Shown below are examples of solutions for other ranges of numbers, in a form that shows chain length, the chain's maximum value, and solution(s). C2 10 [(2, 3), (2, 5)] C3 21 [(2, 5), (2, 7), (3, 7)] C3 21 [(2, ...


8

I believe this is the solution :) Some information on how I solved it:


6

I think it's a safe assumption that the answer will be of the form: The reasoning is that Now, we want to get the longest sequence. To achieve this For instance, if we try Let's try another: Going even lower is not an alternative, since: Without a rigorous proof I'm going to say the combination is:


6

First of all, And now the point is that which means that


1

I got the same answer as hexomino except the right sudoku is so I guess there are at least 2 answers. I like this puzzle idea though!


11

I think this is the answer Partial Reasoning


1

Let's assume for the moment that the judge chooses the starting location at random uniformly, and similarly the direction of travel. In that case, the location at either end of the row is the best for the Janets. Clearly every one of them is equally likely to be the first to be examined. The ones at the ends have only one neighbour, while all the rest have ...


3

First notes: Now, Therefore, That leaves only So the answer is


3

This is only solvable in the sense that no solution is better than another. Since the judge has no prior knowledge, starting at any position is just as good, so the judge might as well toss a (38-sided) coin to decide, (treating the line as if its ends were connected to each other) and this is still optimal for the judge. If the judge does that, it becomes ...


3

After looking at @hexomino's answer I came up with a slightly different proof.


0

1 hour (or less). As soon as the men start digging there is a hole in the ground, so all you need is for each man to start digging their own hole and your done. It may not be as deep as the hole that the 9 men dug when they spent 9 hours digging, but it's still a hole. The puzzle doesn't specify that the new holes have to be as deep as the old ones.


9

Solved grid: Solving path for right grid:


4

This mistake lies:


2

I claim that the long term average number of cars for $N$ starting cars will be To prove this we can proceed by induction


1

(NB: table of calculations below is unspoilered. Smart people, feel free to edit.) Initial remarks N=3 Tedious calculations 4 CARS Back: (1/4*0/3+1/4*3/3+1/4*3/3+1/4*3/3) 2nd-back: (1/4*0/3+1/4*2/3+1/4*3/3+1/4*3/3) 2nd-front: (1/4*0/3+1/4*1/3+1/4*2/3+1/4*3/3) Front: 0 5 CARS Back: (1/5*0/4+1/5*4/4+1/5*4/4+1/5*4/4+1/5*4/4) 2nd-back: (1/5*...


2

Instead of a cartesian plane, let's consider instead a... That means rotating $n$, $m$ and $k$ around, so that So armed with this nomenclature, let's run a few iterations starting with only the $(0,0)$ point: ...we can conclude that the number of moves needed to clear the... ...row/diagonal is...


-1

I believe the answer for part a is one hour, as it takes 9 men take 9 hours to dig 9 holes, then the 9 men dig one hole per hour (9 holes/9 hours = 1), and cosequently one man dig 1/9 hole per hour (1 hole / 9 men = 1/9). Now if there are 3 men then they dig 1/3 of a hole per hour (3 * 1/9 = 1/3) and they are going to work for 3 hours then they dig one hole (...


5

Because which means To be pedantic, however,


2

The answer is Unfortunately, this answer is neither elegant nor easy to explain since I found it via brute force. It's pretty disappointing to solve a puzzle this way, but I don't think anyone explained a correct answer yet (at least before I was sniped by Charles Gleason!). The General Approach Consider the case of 9 coins with one heavier than the rest....


1

A can balance the crates with the following strategy, with or without B's cooperation. Proof: Claim: When A and B are alternating turns, A can guarantee the difference between the crates to always be 1 after A's move. After B's move, the difference between the crates will not exceed 4. Base case: starting with 0 coins in each in each case, A will split ...


8

This can be done in Steps


4

An alternate approach:


1

Let's start by assuming, from player A's point of view, that player B will always place more coins in the lighter box than the heavier one. Later I will show that this assumption is not necessary.


1

With the assumption that you can distinguish the coins by either marking them or retrieving them in the reverse order by stacking them on the scale, then worst case is seven weighings: Arrange the coins into a grid of 4 rows and 5 columns Weigh Row1 against Row2 Weigh Row3 against Row4 At this point, you know either: Which row has the heavy coin AND ...


4

The train is travelling at: Working:


0

Didn't look at any answers: Or am I missing something?


3

Reasoning:


8

I think it can be done in: Method:


4

This can be done in Steps


0

You make four stacks of 5 coins each. You put one stack on one side and one on the other side. You have two possibilities. a) They balance the scale. b) They do not balance the scale. If the scale is balanced then you have the smallest number of steps. If the scale is not balanced, we have the following combinations 9[10-11-10]. Remove the stack with the 9 ...


2

Answer Steps If there is exactly one imbalance $a<b$, then coin $a$ weighs 9 g and coin $b$ weighs 11 g. If there are two imbalances $a<b$ and $c<d$, then weigh only coins $a$ and >!$c$ (one extra weighing) Case 1: $a<c \implies$ coin $a$ weighs 9 g and coin $d$ weighs 11 g Case 2: $c<a \implies$ coin $c$ weighs 9 g and coin $b$ weighs 11 g ...


-1

Weigh two random coins. Luckily, one coin is heavier than the other. Could be: 11 g and 10 g 10 g and 9 g 11 g and 9 g Take both coins (just weighed) and put them on same side (keeping track of the heavier coin) and weigh them against another two coins. You get lucky and both sides weigh the same. If if was the 11 g and 10 g on one side this is ...


8

Here is a new approach. The old solution was not perfect. Logician A can force the crates to balance.


3

A and B have two logical choices. Let's say A choose strategy 1. Then, when it's A's turn again, either the crates will be the same weight, in which case A and B chose the same strategy. If they're not the same weight, that means the B chose strategy 2. If that happens, then A will now change their approach to strategy 2. After that it's straight ...


4

When B is down to three coins left, they are left between the crates. This will happen before A runs out of coins. So the final steps are up to A to carry out. A will use the 3 coins left by B in the same way as A's own coins. Following exactly the same method as before, ensuring both crates have equal coins is simple. The only possible objection here is ...


3

The (twice) given geometrical answer is very clever. Here's a much less clever solution, using very basic engineering only. It utilises the fact that the diagonal happens to be the longest dimension of the brick. Stand two of the bricks up on the ground with some room in between. Keeping one corner always touching one brick, Now you can easily measure ...


3

The answer to this puzzle is: First, focus on the border of nine sets of black/grey squares separated by bars. This is: Next, turn your attention to the eight central blocks of nine black-and-white 'pixels'. Since there is exactly one white pixel in each column we can infer that these are: However, there is a slight twist here... To find the solution, we ...


0

The 7 on top is even further restricted. Every pattern 1-2-1 downwards helps you to exclude. For example it can not be to the left because of 1-2-1 starting in 4th column. Meaning the three in third row must be left. Also the 8 in fifth row has at least one field safe. A lot more if the signs to the right are right.


0

When you tag a problem with [mathematics] you should adhere to the standards. So Example Solution: Checks: Reasoning: Do not read things in the question that are not there! Adhere to definitions when tagging!


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