New answers tagged

1

Answer: Proof:


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As an example, say we have numbers 1, 2, 3. Then, we find the sum of the three numbers, which is six. Then, we take the numbers and lay them out in a row. Then, we ask the sum of two slips. Then, we ask for the sum of another two slips. The one with the greater sum has the biggest number in the group. The one with the smaller sum would have the smaller ...


3

The answers for $M$ are somewhat surprising! I expected smaller numbers. $k=0$ Minimum Maximum $k=1$ Minimum Maximum $k=2$ Minimum Maximum


5

We know that: It then follows that So the question boils down to finding The solutions are therefore:


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This is not a full answer, but it looks like you're getting quite desperate for some response to this, so I'll write up the thoughts I've had so far, in the hope that either someone can take it and continue to get the full answer, or at least that you will see which parts are clear to potential solvers and which parts may still need hints. Basically we're ...


4

First 2 pairs, 69-12=+57, 961-1018=-57 So, the last 2 pairs would probably follow the same reasoning to get +57 and -57 respectively. First ?=2812, Second ?=5269


6

The following arrangement works: where you should imagine the four numbers in the middle on four triangular faces meeting at a vertex, and then the ones around the outside being on the remaining faces adjacent to those.


13

You can get at most This is possible to achieve: So now we've reduced the problem to the variant where Here's an optimal strategy: This is optimal because:


4

Without restriction of only checking consecutive papers The general solution for $n$ would be: Visualization: With condition of only checking consecutive papers I believe this is the general equation for $n$: For $n > 2$:


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Maybe I missed the point here but surely if you used values 1,2,4,8,16....32768 you could tell which two values were selected in any pair, so you'd have at most 2 possibilities for each piece of paper. By mutual exclusion and 'walking' your way along you could identify each value. This would take 14 goes at most.


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Now, when I look at the picture the floor seems to be made of square tiles. It is also said that they are in the middle of the room. So, they just have to feel the floor for the crevices made by the stones on the floor(close to their position). When they find two corners on the floor, they have to go in the direction facing the center of the two ...


1

I think value of x is : Reasoning(for first puzzle) : Reasoning(for second puzzle) :


2

I'm not 100% sure here but my pattern seems to work. The value of x is Explanation : I might have entirely missed the pattern and found a wrong answer, but this is the best I could come up with.


4

I can prove that is minimal. The difficulty comes in distinguishing each of the 14 pairs of neighboring slips of paper, as well as the first and last papers. There are 15 such pairs of slips of paper. If a question includes or excludes both slips, the response will be the same if those slips were swapped. In contrast, a question can only distinguish the ...


2

Exact probabilities, in case anyone is interested.


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A simulation of 100,000,000 of such best-of-seven series of A:B showed ... And that's the code for simulating: import random from collections import Counter def sim(home, away): i = random.randint(1,100) if i <= 51: return home return away def series(): w = [0,0] for game in range(7): if game <= 3: # fix: < ...


11

By choosing appropriate numbers to write at the start, you can manage it with Here's how. Choice of numbers The numbers you should write at the beginning are Let's label the numbers, in the new order as they're laid out face down, $a,b,c,d,e,f,g,h,i,j,k,l,m,n,o$. Every time you ask a question about some subset of consecutive papers, you know Note that ...


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This is simpler than it looks: no calculations are involved. The trick is to realise that being the first team to win four games is Then,


2

The answer is Reason


14

First of all, let's see why your brute-forcing fails. (This is the puzzle part, the rest is plain old math.) No matter which you chose, the number at the bottom right would have to be both odd and even at the same time, so there's no integer solution. However, there are four equations and four unknowns, so we should have at least one solution (unless the ...


1

Travel time The real danger Taking the black? Conclusion


10

The one who is guaranteed to die before they reach the Wall is: Because:


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If this is allowed you could ask him to evaluate the polynomial at $\pi$ or another transcendental number. The response will be something like $a_0 + a_1 \pi + a_2 \pi^2 + ...$, which is unique for all combinations of $\{a_i\}$, even for negative integers (by the definiton of transcendental numbers). The drawback of this method is just that depending on the ...


3

It's not possible to find it with a single query. Suppose you had one, e.g. asking for a single number $n$. Then you can't distinguish between the cases where your friend's secret polynomial is $P(x) = 2x$ and $P(x) = x + n$; in both cases, they will reply with $2n$. (I know it's unlikely that he chose the second option, especially if $n$ is a non-obvious ...


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1

It follows So,


2

6 = As the sequence is


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Well, judging from the hint, each $[m\ast n]$ means In our case, we seem to have only $[n\ast n]$, in which case So rewriting all the expressions we've been given: Word 1 (2 letters): $[3*3]+[1*1]+[2*2]+[2*2]$ Word 2 (5 letters): $((-[1*1] + [3*3] + [4*4] + [8*8] + [27*27]) \cdot [2*2] + [1*1] + [1*1] + [5*5] \cdot [27*27]) \cdot [2*2]$ Word 3 (2 ...


3

The answer is most likely Because


6

Using only basic operators (+ - * / ^) There are 16 solutions that, like yours, are missing the center column: There are 64 boards total that are missing any one row, column, or diagonal. Here is the complete list: https://pastebin.com/dSAnwQhB and here is the generating code: https://repl.it/repls/BriefWarmInstitutions If we were to introduce ...


2

I used integer linear programming to verify that the problem is infeasible. It turns out that there are 64 near misses with one violation like yours. These reduce to 8 solutions up to symmetry:


5

This is obviously not the answer you intended Examples: Therefore:


6

Bonus answer edited in after initial: So!


1

Full Answer except for $51,67,68$


0

Partial answer


3

Partial answer (all except 28, updated after .2 was allowed)


5

All 1-50 solved. Solved but extending meanings of operators: $28, 34, 38, 42, 43, 44, 45, 46, 47, 49, 50$


1

Following the rules from last time:


10

Overall solution The nine symbols are, in the arrangement given: Step-by-step deductions From the factorial relationship, But from the perfect-power relationship, So The top left (division) relationship From the square root relationship, Going back to the perfect-power relationship, Now for the big product relationship: From the division ...


8

The answer is The solution method is: Cryptography of the day: And to answer the question:


4

The rule is: Question 1: Question 2: Question 3:


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"Can the two logicians redeem themselves? If so, what will the reasoning behind the correct answer be, and what's the minimum number of days it will take either of them to answer correctly?" [NOTE: This is flawed at this moment, I'll remove this note if I can fix it.] Answer: Main Line of Logical Reasoning: [This all occurs before any passing or answer.] ...


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My initial answer (without a proof): Click this link to see the (bad) code. Edit: The first version of the code did not take into account draws and it allowed games to continue after a player has won. After fixing these problems: These results come from players making thoughtless choices and are not representative of perfect games. In order to ...


16

The rule is: The missing 7 digits are:


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“Is this method general, meaning could you do it with any number of bags?” Yes, it’s possible with any number of bags depending on the amount of coins in each bag. If you want to investigated N bags, using the method described in Columbo (thus: only 1 bag with fake coins and taking 1 coin from the first bag, 2 coins from the second etc.), the following ...


1

I think the most straightforward pictorial description is to have a picture for each nail that is coloured segment of a disc. Any combination of nails for which the pictured segments can be put together to exactly form a disc of a single colour should release the string. So the puzzles would be: Puzzle 1 (1-out-of-3) Three discs Puzzle 2 (2-out-of-3) ...


2

My concern with this whole approach is that you're trying to reduce to abstract figures some instructions that are fairly easy to explain textually but may prove somewhat more challenging to unambiguously depict visually, especially to children. I would guess the objective is to reduce the instructions to this visual shorthand and then write/etch/whatever ...


1

My idea for how to visualize this is to depict the nails as having interior regions which are partly filled and partly unfilled, with the conceit that whenever all of the removed nails, when overlapped with each other, together have a filled portion which fully covers the entire interior, the picture will drop. I will demonstrate this for Puzzle 11: As you ...


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