# Tag Info

20

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

16

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;

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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

7

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?

7

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

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Immediately, $C, I, L, U, V =$ Equations:

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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 ...

5

Just a guess, but...

5

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.

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. ...

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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 ...

4

I guess the answer is Because And to solve this Alternatively from @hexomino

4

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One possible solution Reasoning Alternative,

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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 ...

3

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

3

Here is one solution. Some observations

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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..

3

The Missing Members may be If this is a countdown of

3

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

3

Using the Vinculum: we have:

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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 ...

2

Here's a solution:

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I have managed to get better than the existing answers for a maximum distance from Savage to Hogan of With the following sign choice I think the actual maximum is within 10 of this as I've been forced to use all the 1s and 2s at this point.

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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:

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Solution Reasoning

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This may be far too simple but: Reasoning:

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I think it's

1

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