# Tag Info

1

Since only a few combinations can work off the bat. For example, But this one doesn't work. Looking at more that do, I found Which is the answer.

3

Okay I think this is the solution

3

Edit: a revised solution Some reasoning: My first solution

1

I believe the following values are consistent with the equations: My first discovery is that Working from Equations 1 and 4: This gives us: Now back to equation 2: Finally with equation 3:

7

It's the directions you run when the cops are chasing you, a.k.a. The way you get this: (Sorry if my formatting sucks, I'm still new to this site!)

5

It's: because:

1

ma is the age of Mark my ist the age of Mindy

2

Let a be Mark and b be Mindy. Then So

1

Assuming that when the final card of the deck is drawn, the running total remains and the deck is reshuffled. The game will continue at most 3 times through the deck. The count begins at zero (0 modulo 3), after the first pass through the deck the sum is ten (1 modulo 3), and after the second it will be twenty (2 modulo 3), and then thirty (0 modulo 3). ...

1

Wrote an R script: install.packages("combinat") require(combinat) #Creates all permutations for drawing four cards data<-permn(1:4) table<-matrix(nrow=length(data),ncol=4) table<-as.data.frame(table) for(i in 1:length(data)){ for(j in 1:4){ table[i,j]<-data[[i]][j] } } #Don't care about Jill's second pull table$V4<-NULL table$sum<-...

3

I'm going to assume that if the fourth card is drawn, and a multiple of 3 is never reached, the game is considered drawn and this is not a win condition for Jack. Let's start with the easy part: Now going through other possibilities: If Jack draws 1: If Jack draws 2: If Jack draws 4: Ultimately, this gives him odds of: EDIT: Adjusting for the fact ...

1

Edit: The maximum distance marker I have managed to construct is Using the following placement of stickers (in bold as suggested). Progression on the upper bound Original I had originally thought I had a solution with distance Using the following signs But as Weather Vane correctly pointed out in the comments, I had constructed a new sign using only ...

4

4

Using the Vinculum: we have:

3

This may be far too simple but: Reasoning:

0

Disclaimer: I'm typing this out as I'm thinking this through, so it's a bit disorganized. I'll go back and work on cleaning this up when I get the time. Initially, looking at the rings: Further testing of these combinations thins this down to: Looking at the rings: Likewise, Now working from Ring E: Finally: Therefore we come to the final solution of:

2

Solution Reasoning

0

The problem can be reduced to the following scenario: where:

1

Simple proof of impossibility:

1

Well, this one is much easier than the last one, with more extra clues. I did all this without looking at either Daniel Duque's answer or my previous partial answer. Step-by-step deduction Now we have: Now we have: Done! Final solution

6

Found a solution requiring First, label the coins A,B,C,D,E,F,G,H,I,J. Then for the first 4 weighings, weigh: For: For (Update: fixed mistake, thanks to comments): For: For: In any case, And as has been pointed out, there are 360 initial configurations, and $360 > 243 = 3^5$, so at least 6 weighings are required.

3

I am not entirely sure about what the OP means by needing knowledge from previous puzzles to understand the meaning behing the colours; nevertheless I figured a path the knight can follow though his realm: Unfortunately there is not much I can add about how I got there, and it would be very extensive to explain all 64 squares:

17

We have the following COCA +COLA ----- SODA Next, notice something similar in the Since we have a 4 digit number as the result, we know that But: Thus, Also, we know Thus, the solution is;

3

This is a depiction of numbers in ternary (base 3). Effectively your right hand column represents 'ones', the middle column 'threes', and the left hand one 'nines' (in the same way that in decimal we have 1s, then 10s, 100s, etc.). A green square implies a value of 0 in that position, red a 1 and blue a 2. e.g. 10 red-green-red = 101 in ternary, i.e. one 9 ...

13

8

Since we know that Therefore $A$ Hundreds value must carry since $O \neq 0$ Therefore Therefore $O$ We now get And since $S<9$ Then there are many possibilities... any relations I missed out?

0

I think 6 will workout.Let us divide the total coins in set of 3,3,3,1. Now when you compare 3(1),3(2)/ 3(1),3(3)/3(2)3(3) ypu will get various weights increasing order. I will use mathematical forms to represent the possible weights. Possibilities are +,0,- +,+,0 ++,0,- +,+,- 0,0,+ By internal comparison you will be able to get in 6 minimum chances.

3

One possible solution Reasoning Alternative,

3

Step-by-step deductions (partial) Filled grid (partial) (A big number means that number is definitely in that square. Two or three small numbers means that square must contain one of those numbers: e.g. G1 must be either 33 or 63, although either 33 or 63 might also be elsewhere. A number with a question mark means that square is one of only a few ...

7

Full Solution Notation: Deducing $e$ therefore, Deducing $c, f, h$ regarding B, C: only pair available for distance of 4 is: Therefore, Sum of 25 rule renders Deducing the rest regarding A, E: only pair available for distance of 2 is: Therefore, Sum of 25 renders

5

TL;DR Preliminary deductions Left column: Bottom row: In particular, the numbers Option 1 Let's assume Top row: Right column: So we have The remaining numbers are Option 2 Let's assume Top row: Right column: So we have one of the following two possibilities: The remaining numbers are respectively $2,6,8$ or $4,6,8$, so the complete grid is ...

4

-1

thief ran one of two known directions and the cop is faster than the robber So it's garanteed if the cop goes in the known direction for some finite amount of time. Life is guaranteed by no one, so we can't garentee he'll catch the robber.

3

It works with We can rewrite the equation as follows: Clearly this works if which yields the solution stated.

1

1

Not sure if this is optimal, but with my strategy I need at most balancing actions. We use the following notation: Here is the tactic: There is no way that this step takes fewer turns using this tactic because: Now that we figured out which case we have, we can do this: We are done.

4

I have a method that will identify the fakes in weighings: Start by There are three possible outcomes: Let's handle these separately, starting with the easiest case that has Next, let's handle the case with And finally, there could be Now, if only we were able to shave off one weighing from the "only 1 unbalanced" case..

5

Not a solution but the theoretical minimum weighings required is 6. There are 360 combinations of coins. 10 locations for the smaller fake and then 2/9 locations for the larger fakes: 10*nchoosek(9,2) = 360. Each weighing provides a ternary bit of information. At a minimum, 6 weighings are sufficient to learn all fake coin locations: 3^6 = 729 > 360. ...

1

I think it's

1

Lets give it a try. We name the coins $HHLGGGGGGG$ (Heavier Lighter, Genuine) Weighing 1 Weighing 2-5 After making Weighings 2-5 you have a pattern with 4 results H (Higher), L (Lower), E (Equal), ordering is not important Now, if your Pattern is ... So go on with what is left Solution

3

Here is one solution. Some observations

20

Yes, it's possible. But what if It's still possible in this case:

5

Just a guess, but...

2

Answer in construction From an algorithm point of view, the ways you can attempt to crack the problem are either too strenuous for decent results in an amount of time or redundant in comparison to a decent mathematician... so I don't expect anything useful to be yielded from using a computer algorithm. That said, I have a simple starting point that may or ...

7

I think that in the first case the total sum is Reasoning For the second case

6

Absurdly partial answer Terminology: an "inner configuration" is an arrangement of filled and empty cells in our $n\times n$ grid; an "outer configuration" is the corresponding arrangement of run-lengths displayed outside the grid. Let $I_n=2^{n^2}$ be the number of possible inner configurations, $O_n$ be the number of distinct outer configurations arising ...

0

I doubt it could be calculated exactly except for very small grids, but it should be possible to Monte Carlo it. One has to be careful about bias. I don't think there is any way to choose the numbers directly such that all solvable number patterns are equally likely. Instead you could choose a random grid picture such that every picture is equally likely (...

3

I've found a solution, which is unique given the constraints. Proof

2

Here's a solution:

3

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