# Tag Info

## New answers tagged logical-deduction

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

0

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). ...

0

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<-...

2

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 ...

6

The first is a pirates favorite letter. The second is from a famous quote of Shakespeare. The third is squaring it off in a famous formula. The fourth is the first in a very well known song. The fifth placed in a line from top to bottom makes the train go in the right direction. The sixth in my language sounds like the fourth. The seventh is what you ...

4

Using the Vinculum: we have:

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

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;

13

Based on Omega Krypton's answer,

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?

2

A completely sideways answer - but possibly only one try... Here you need to cheat a little, and cause yourself some sort of injury to get some blood - smear some of that blood on one card, and then another - keep doing this until you find two that have different shades. The red "filter" cannot restore invisible (to you) colour, but it can hide some of the ...

3

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

2

Here's a solution:

0

Just for a bit of variety same answer as above but using PL/SQL ABCD EFGH IJKL MNOP ------------------------------------------------ AEIM BFJN CGKO DHLP ------------------------------------------------ AFKP BELO CHIN DGJM ------------------------------------------------ AGLN BHKM CEJP DFIO ------------------------------------------------ AHJO BGIP CFLM DEKN ...

0

I can do it in

12

I can prove a lower bound of The first step is to notice that the question is equivalent to The reason is: and Then we observe and So there are a few cases. If If If

1

Exactly the same result as Markus Jarderot, who got there while I was writing.

5

ABCD EFGH IJKL MNOP AEIM BFJN CGKO DHLP AFKP BELO CHIN DGJM AGLN BHKM CEJP DFIO AHJO BGIP CFLM DEKN Generated with C#: void Main() { // List of players A...P var players = string.Join("", Enumerable.Range(0, 16).Select(i => (char)('A' + i))); // Set of players each has yet of meet. var unmet = players.ToDictionary(c => c, c => ...

13

Here is one way to win almost all the candies: I think this solution is optimal, in the sense that you cannot end with more candies.

23

It can be done in Solution

16

Also not sure whether this is optimal but I can do this in First, In the worst case, Finally,

1

Not sure if this is optimal, but: Then worst case, after So you're guaranteed to find two green on your

1

From the hint, since we have and then it follows that Hence both Now since and we already know we have Hence All the other equations are superfluous.

-2

If I can get the cards back after an attempt Otherwise,

11

This is like a reverse puzzle to the well-known Here in reverse: A remark as a thinking outside the box solution:

13

I believe I can open it in The reasoning behind it is:

-1

Poison Wine Forward Poison Poison Wine Backward Explanation CLUE: Fourth, the second left and the second on the right Are twins once you taste them, though different at first sight CLUE: Two among our number hold only nettle wine, Therefore bottles 2 & 6 must be nettle wine as they are twins and there are two wine. CLUE: First, however slyly the ...

3

My attempts: From (iv) and (v), From the hints, Verification

5

Size considerations (starting from the left) Modular considerations (starting from the right) Case checking So far we have: Let's try So Final answer

4

Partial answer that I'm saving for now (For convenience, I will call $PCRON$ "the root" and $PRINCETOM$ "the square". We can first deduce that the digit N We can also do some quick tests to find the approximate range Let's try that: We can then determine that C is

5

Theorem: for any natural number $N$, the expression is a perfect square. Proof: by induction on $N$. The result is true for $N=1$, since Assume the result is true for $N=n-1$. Then we have, for $N=n$, The extra factor there is which is a perfect square. Thus the theorem is proved. So we can remove Can we do any better? So if we remove just one factor,...

0

Some thoughts

-3

strong textOnly factorial that would be perfect Square is 1! Therefore, from 2! To 2019! Have to be removed.

16

Rather remarkably, I wrote down this exact puzzle in my notebook a couple of years ago to which I think the answer is Reasoning

4

I believe A, B, P are Some deductions I came to before brute forcing

2

I agree with @w l on part 1 of this question, we all seemed to get the exact same pattern. Part 2 is still open, so here's my take: Also

4

I will assume that no one can make nonsense statements.

3

Solutions (all names here and below are abbreviated to initial letters): Explanation Verification

1

Assuming the magician's guesses involve revealing the contents of the cups he guesses, correctly guessing both the cups that hold a ball within four guesses can be accomplished with a rather simple strategy. As such, the magician can always guess the two correct cups with at least four guesses as long as the contents of all guessed cups are revealed.

3

I think I found more possible solutions. All of them are compatible with the puzzle, though it wouldn't make a lot of sense for Kaito to ask for the number, yet.

1

I'm pretty sure this is similar to some other answers, but it's presented in a way that's more practical to the magician and their assistant. When the assistant indicates the cup, The heuristic for the assistant's selection is: To be exhaustive, all the permutations are:

7

The Answer is: I just fixed the first picture and arranged the others according to it.

1

The apprentice arranges the cups such the balls are amongst the first four cups closest to the doorway. The apprentice allows the magician to enter the room. The apprentice says, 'Come in and pick the from the four cups closest to you' This also works for a two guess strategy.

4

Second solution with each symbol matched. I wasn't happy with my first solution because the red stars are not oriented the same. This is simpler and more pleasing: My solution

5

9

I think the answer is Reasoning

5

Thinking completely sideways, the assistant picks a cup (won't matter which one, just the actions) put hand on cup. wait n seconds. lift cup. wait m seconds. drop cup. ball 1 is n cups away, ball 2 is m cups away.

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