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Weighing Coins -- a Different Approach

We know that $3$ weighings suffice for $13$ coins if we have one known good (see here), so for $4$ we can split $39$ coins into three groups of $13$ and weigh two of the groups against each other. ...
Ross Millikan's user avatar
0 votes

5 guards, and get rid of the timer!

So first, I don't believe it is possible to guarantee a solution in 5 questions. The reason is because you always have a 20% chance of your first question going to D, which makes any conclusions you ...
Truej's user avatar
  • 532
0 votes

Weighing Coins -- a Different Approach

I would assume that for 4 weighings you could figure out the false coin for n<=27=3^3 without requiring some sort of luck. When you know whether the single coin is heavier or lighter than the rest,...
patrooon's user avatar
0 votes

Green Eyed Oracle Variant - Not a Multiple of 17

A bit of a partial answer,
uggupuggu's user avatar
  • 163
3 votes
Accepted

Identify the truth teller, the liar and the spy

NoeS's user avatar
  • 324
0 votes

Knights, Knaves, and Spies

As others have stated, A = knave, B = spy, C = Knight is correct. Proof using linear programming and Pyomo software within Python. Sets: $I = (1, 2, 3)$ where (1,2,3) represent the three given ...
Faisal Ahmed's user avatar
8 votes

Knights, Knaves, and Spies

Your belief that, A = knave, B = spy, C = Knight is correct. In knight/knave/spy puzzles, it is often best to postpone figuring out who the spy is because a spy can say anything. In other words, a ...
Will.Octagon.Gibson's user avatar
5 votes
Accepted

Colombian Sudoku, Cows and Bulls again…

Completed grid: As usual, we begin by using our given digits to mark cells which match and cells which cannot match. We then: As usual, I forget that the lack of a clue in column 8 actually means ...
Jeremy Dover's user avatar
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1 vote

Knights, Knaves, and Spies

B and C are in agreement, so either both are telling the truth (one is knight and other is spy) or both are lying (one is knave and other is spy). If B is telling the truth: "I am a spy" ...
lnx's user avatar
  • 311
1 vote

Largest prime number erroneously erased when finding all primes less than 1000

I got the answer after some thinking. When you erase a prime number, the square of the prime number escapes from being erased. For example, if you erase 3, 9 will escape. But when we erase two prime ...
Rahul Krishna's user avatar
0 votes

Eliminate some numbers so that each of the three rows contains the numbers 1 through 9 each exactly once

...
catpop's user avatar
  • 1
1 vote
Accepted

What is the name of this puzzle type? How to create one? nxn grid where you select n cells with different numbers

The solution is called a transversal in an $n \times n$ array with $n$ symbols (for example, in a Latin square): https://en.wikipedia.org/wiki/Latin_square#Transversals_and_rainbow_matchings For a ...
RobPratt's user avatar
  • 14.3k
3 votes

Fill the triangular grid using the digits 1-9 subject to the constraints provided

Gave it a try in "hard mode", where the two knowledge-ish diagonals (MCCCXCVIII and Fahrenheit) are not used in the solving process. It turns out that it is indeed possible to find the ...
Bubbler's user avatar
  • 16.3k
5 votes

Fill the triangular grid using the digits 1-9 subject to the constraints provided

Solution: Solving Path
Lukas Rotter's user avatar
4 votes

Eliminate some numbers so that each of the three rows contains the numbers 1 through 9 each exactly once

Lucenaposition's user avatar
8 votes
Accepted

Eliminate some numbers so that each of the three rows contains the numbers 1 through 9 each exactly once

So the starting combination is ...
Novarg's user avatar
  • 4,057
24 votes
Accepted

Are you radical enough to solve this SURDOKU?

Surd-lution (sorry :P): To start: Now the step by step: 1: 2: 3: 4: 5: 6:
Beastly Gerbil's user avatar
14 votes
Accepted

Fill the grid subject to product, sum and knight move constraints

Bubbler's user avatar
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4 votes

Fill the grid subject to product, sum and knight move constraints

You can solve the problem via integer linear programming as follows. For $i,j\in\{1,\dots,5\}$ and $k\in\{1,\dots,20\}$, let binary decision variable $x_{ijk}$ indicate whether cell $(i,j)$ takes ...
RobPratt's user avatar
  • 14.3k
7 votes

Fill the grid subject to product, sum and knight move constraints

First find a place for 19: Then for 17: Then for 18: For 11, 12, 13: For 14, 15, 16: Continuing: The completed grid:
Daniel Mathias's user avatar
4 votes
Accepted

Next date in the future that satisfies the following conditions

This unpolished script: yields all the upcoming dates satisfying the three conditions. The first of these is The first that also satisfies the bonus criterion (equivalent to being in a year ...
AxiomaticSystem's user avatar
4 votes
Accepted

A colourful Colombian Sudoku

Surprisingly very tricky! Here's the solution: Step by step: 1 - First steps: 2 - Completed clues: 3 - Cross checking: 4 - First numbers: 5 - It was all yellow: 6 - Contradiction: 7 - Little ...
Beastly Gerbil's user avatar
3 votes

Find the value of '?' in 4x4 table of numbers

The entry is, or rather may be $85$, with the following way to guess it. Take the first and the third columns. Add them. To the result add four times the second column. We obtain: $$ \begin{bmatrix} ...
dan_fulea's user avatar
  • 352
10 votes
Accepted

Tombs of Ancients

Gladys is visiting Solution Step-by-step
user39583's user avatar
  • 9,103
17 votes
Accepted

Next date in the future such that all 8 digits of MM/DD/YYYY are all different and the product of MM, DD and YY is equal to YYYY

I believe the first such future date (represented as MM/DD/YYYY) is From the answer to the previous problem
hexomino's user avatar
  • 138k
-2 votes

Prove that all 101 balls have the same weight

If we first suppose that your initial number of pounds for each ball is the same, the total weight of all 101 balls is equal to: 101 * w, w being the weight that each individual ball matches. If your ...
Omar's user avatar
  • 1
1 vote

Prove that all 101 balls have the same weight

Here is an attempt to explain the provided solution more simply. The structure of the problem depends highly on odd-even (parity) arguments, but this provides a new way of looking at it, that is ...
dan_fulea's user avatar
  • 352
9 votes
Accepted

What should the next step be in this Kakuro puzzle?

There is one possible deduction: But it ends there. Using a computer program, I confirmed that every candidate in every cell in the image below is part of some potential solution. There are 24 ...
Bubbler's user avatar
  • 16.3k
6 votes
Accepted

Bulls and Cows meet Sudoku!

The final grid: Starting with matching, we can: Initial analysis: A little bit more subtle: And a little less subtle: Let's do some sudoku: Some more match logic: Seeking 9: Ah...column 2: ...
Jeremy Dover's user avatar
  • 28.8k
1 vote
Accepted

Lewis Carroll's Logic Problem: what do you know about great-grandsons?

I coded this in Fortran to search for combinations of pairs of persons along with their relationships that satisfied all problem requirements. The simplest family tree involves 6 people but there are ...
Bob Bixler's user avatar
3 votes
Accepted

Finding a mystery number - with three logicians

To write down the solution, i need some notations. There are three perfect logicians involved, Adam, Beatrix, and Caesar, but since the solution will be doubled by code, $0,1,2$ in this order may be ...
dan_fulea's user avatar
  • 352
0 votes

Finding a mystery number - with three logicians

Rewriting an answer for this. The logic ends up pretty hairy, and it doesn't seem like the answer is unique. EDIT: with the change, there is a trivial final step. After A's statement: After B's ...
kagami's user avatar
  • 824
0 votes
Accepted

Justifying a redouble

Here is the intended solution:
Parcly Taxel's user avatar
  • 7,820
0 votes

Lewis Carroll's Logic Problem: what do you know about great-grandsons?

Fact: Using: We can rewrite: Then using: There is also the trivial:
gannolloy's user avatar
  • 1,600

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