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;
12
Based on Omega Krypton's answer,
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
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
6
Immediately,
$C, I, L, U, V =$
Equations:
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 ...
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.
...
5
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
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 ...
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
3
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:
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 ...
2
Here's a solution:
2
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.
2
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:
1
This may be far too simple but:
Reasoning:
1
I think it's
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