LOGIC AS USED TO SOLVE THE PUZZLE: The first thing to note is that with 3 points for a win, 1 for a draw, and the known points totals, we must have the following end outcomes in terms of wins, draw and losses:
In particular, for teams B and C either scenarios (a) would both happen together or scenarios (b) would both happen together (since we need an equal ...
Given the wording of "whose turn it is", we are left to assume that taking out the trash will be a recurring event. In this way, we can guarantee fairness regardless of $n$ and $p_i$ by simply recording who did it last and alternating each time.
The coin flipping only needs to occur for the very first time, but on a sufficiently large timeline, one sample ...
A set of coins is fair in the relevant sense if and only if
Proof (slightly highbrow, sorry):
Alternative kinda-equivalent proof (simpler ideas but needs you to know a theorem):
(Neil W suggested, in comments, taking that second approach. I'd avoided it because the other way seemed quicker and more first-principles-y, but the second way may well be easier ...
Since Hehe knows his own number by watching the others, one can test the different factors that could lead to it (watched hats, with ascending sort -> possible hats Hehe could have):
So we only keep some of these entries:
And just consider:
Hehe knows his number even before the other ones talk. This means: a-priori he knows, with no ambiguity, the number ...
I took a slightly different approach to Jaap Scherphuis:
The two smaller numbers are factors of the largest number, and all 3 numbers are different. This means that the largest number cannot be prime - ruling out 1,2,3,5,7, and leaving us with 4, 6, 8, and 9.
There are 32 possible pairs of factors of the numbers 4, 6, 8, and 9, but only 8 of these ...
What combinations of numbers are possible?
From this list of possibilities, we need to find one where one person can know what their number is by seeing two of the other numbers. Here are the possible numbers a person could see others having (lowest number first), and the possible numbers that the person could be.
Now we have 3 possibilities for two ...
It is not possible to do this in 12 steps. In fact, a simple argument shows that at least 20 steps are necessary:
Tiles 1 and 8 need to be swapped. They are however a distance of 3 moves apart, so each tile needs to be moved at least 3 times. Similarly, tiles 2 and 7 also each need 3 moves to reach their goal locations. So these 4 tiles together need at ...
This was trickier than it looked. I have a feeling that there should be a quicker way to find the answer than the list of cases I worked through.
Note: I interpret "two of which are factors of the third" to mean that two of the numbers divide the third, and that either of those two factors might be 1.
Say the starting point is row number $n>1$, column letter $l$, and exactly one cell is omitted. The ant's journey can be described as follows:
The only possible starting point is
With the following code let prolog search the results - X, Y, Z are the coefficients for each bag.
list_allperms(L, Ps) :- bagof(P, permutation(L,P), Ps).
f(X,Y,Z,R) :- R #= 10 * X + 11 * Y + 12 * Z.
fit(X,Y,Z) :- f(X,Y,Z,R), R #< 51.
fit_limit(L) :- nth0(0,L,X),nth0(1,L,Y),nth0(2,L,Z),fit(X,Y,Z).
fit_unique(L,R) :- nth0(...
As JNF says, your mother must have AB parents and be BB herself.
Applying the final point directly to your father, before taking into account information from your own blood type your priors for his genotype are
OO: $0.25$, AB: $0.25$, AA: $0.125$, AO: $0.125$, BB: $0.125$, BO: $0.125$.
Thus the probabilities of a given allele from your father (again ...
This answer will explain in detail how we arrived at our initial simple solution.
The question that you can ask the guard is:
The explanation of why it works:
To arrive at the given answer (question that solves the puzzle), we made the following observation:
I originally missed the piece saying that the truthteller holds the key to freedom, so I solved the harder puzzle where that information is not known. I still like this, though.
I ask guard A:
Based on the answer:
Here is the question I came up with when trying to mix 1b and 2a from above:
Let's break this question up a little more.
First, let us look at the case in which the person being asked is the truth-telling guard who happens to be guarding the freedom door and have the key to the freedom door:
This makes this combination SCREECH because both ...
Answer for the first Puzzle - The amount of triangles that are directed up & down should be balanced. 7 stripped triangles UP - 7 stripped triangles DOWN
To make the balance for the blank one you need to choose an answer D.
Here's a question which I think works.
(Note: I came up with this without reading the extra "warming up" section, so my question doesn't produce the exact same table that you had, but can be easily modified to do so.)
That is, a question is needed such that sometimes the guard cannot
answer, and other times he can answer with both a yes or a no (it does
So the question is about time. Meaning sometime you ask the question they may not be able to answer, while other times they can. This means you can ask the same question multiple times.
The Answer is
I don't have a complete answer, but I feel like this way of looking at it makes some progress:
Their answer consists of two different bits of information: whether they can answer 'yes' consistently with their truth inclination, and whether they can answer 'no'. If they can only answer one, that's the answer they give. If both are possible, they screech. ...
Question I'd ask: "What would the other troll tell me is the road to my destination?"
Liar would tell me truth-teller would guide to death.
Truth teller would tell me liar would guide me to death.
I go the on the opposite road and reach my destination.
A brute force solution: We start with the two sixes with six empty places in between. We must fill those six places with exactly one of each number 1 to 5, as well as 7, while following the rules of spacing.
Now, filling in the 3:
When placing the 5, we can easily reject any solution that would leave more than 1 empty place in each suit, since we ...
This works, although I'm not sure if it's unique:
How I found it
We have two 6s (red and black) with six cards between them that must be all the same colour (let's say red). Key fact:
Now, to begin with,
Since most of the information is in point 4, it seems sensible to start the reasoning there.
4(a) If clubs are the trump suit, then both the queen with the fruits and one with the wheat ears can beat the jack holding a scythe.
From this we get that the suits of the fruit and wheat queens must be
Combining this with
The queen of the ...