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# Tag Info

26

I'm not sure if you don't understand the example puzzle, or if you don't understand how to find the solution. I'll explain both, just in case. The example takes a word/phrase that is 13 letters long with no repeated letters. That means it contains exactly half of the letters in the alphabet (26 total). Then the 13 remaining letters are written below, in ...

20

Salon is a news website - "eds" refers to "editors" here. The clue attempts to misdirect you into thinking of workers at a spa by putting the name at the beginning.

17

The most common solution is usually stated as "If I asked the other guard if the left door led to freedom, what would he say?" This makes the chain of statements "run through" the liar once and the truthteller once, so you know the door indicated is the wrong one. ("Yes" means you should go through the right door, "no" means you should go through the ...

16

You need to find some systematic technique. The technique will depend on the problem. Usually there is a regularity to the problem that you can exploit. In this case, I would note that the two squares that are not part of the main $4 \times 4$ do not interact with the lattice, so we can count them separately. For the lattice, work by square size. You ...

15

The answer is: First row: Second row: The elements are:

15

In more an English sense than a Puzzling one... In the question there is the keyword: That means if 100% true, then the statement is true. else it is false, even it is true 99%. In other words: Hope this helps!

15

It seems easy enough to explain how most of this applies to But why in particular? (I've looked in a few places and everyone agrees that this specific thing is the intended answer.) I think there are two clues. The first is in this bit: My frettings loudly rush and ring Above the people and most clearly sing When I forth-fare on air I'll quote a ...

12

The coding phrase is: The resulting codex is: Found by:

12

As you have posed the question, the four conditions clearly don't work. The prisoner can't conclude that he won't be hanged because the judge said he would and the judge is truthful. As a perfect logician he would recognize that. It would be quite tricky to come up with a problem statement that works well. The paradox rests, I believe, on a rather neat ...

12

This is a variation of the Seven Bridges of Königsberg puzzle. The answer is you can't trace out the pattern if more than two nodes are odd. We call a node "odd" if it has an odd number of lines to it, and "even" if an even number. (If a node is not an endpoint of the drawn line, then the drawn line must enter that node the same number of times as it ...

10

It appears to be two clues in one: "Chimpanzee, say" (answer: APE) "Circus equipment has APE at the centre." (The letters "APE" are at the exact centre of the word "TRAPEZE".)

10

Setup We have four people (B,G,K,N), four objects sighted (balloon, kite, plane, telephone pole), and four days (Tuesday, Wednesday, Thursday, Friday). Here's the information we've been provided with: K's day was earlier than balloon day. K's day was later than kite day. G didn't spot the kite. Friday was either B's day or plane day (or both). N's day wasn'...

10

Based on his thought process: "But somebody insist that his profit is just 10 because he had used his first 10 profit to buy back the goat for 80. " But honestly, a better way to see it is that:

9

I am whole but incomplete This means a skeleton is whole considering it as a structure. But it is just a part of the body. So it's incomplete. I have no eyes, yet I see. You can see, and see right through me. This means skeletons don't have eyes - just the holes. By yet I can 'see' it refers to the 'see' in the word 's'k'e'l'e'ton. The same is meant ...

9

Systematic technique is definitely required. I came across this particular puzzle a few years ago on facebook and got into arguments with some people about the answer, with them arguing that there were or even more squares. Eventually I got bored, produced this animation which shows 40 different squares, and challenged them to show me one that I'd not ...

9

Deusovi has already shown why the answer must be what it is, but there's one thing that can still be added to the answer. That is, actually drawing this many points on an actual cup. While I've already done just that (it was years ago), I have no visual proof that I actually did. So, a 3D render will have to do. Here's the same cup without its handle: ...

8

When we are a fraction $f$ of the way through a 12-hour "day" -- so $f=1$ means 12 hours have passed since noon or midnight -- the positions of the hands as fractions of a whole turn are: $f$ (for the hour hand), $12f$ (for the minute hand), and $720f$ (for the second hand), where integer differences are not visible. So for all three to coincide we need the ...

8

I think this is just a matter of understanding the language used in logic. In the implication If A, then B you seem to be arguing that, since there are cases where A is true but B can be either true or false, we should say "the implication is neither true nor false". However, every mathematician I know would say that the implication is false. In order ...

7

How can Albert, with the information he has after the second statement, make the third one? In your post, you have already accepted that Bernard was not told the 14th. This leaves the following numbers: 15, 16, 17, 18, 19. However, since Albert was told "July", before Bernard has even said anything, Albert has eliminated 15, 17, 18, and 19. This leaves only ...

7

This is topologically equivalent to a torus, and you can go up to 7 points: as shown by this Math.SE answer. The diagram for this could look for example like this: One can also just look up the answer if you know the question asks for the complete graph $K_n$ of degree $n$ with maximum $n$ such that the graph genus $\gamma (K_n)$ is at most $1$. Then

6

Yes, you are correct. Those mistakes should be fixed in order to make the questions analogous.

5

Another method of solving: Take the list of letter frequencies in English (usually something like ETAOIN SHRDLU CMFWYP VBGKQJ XZ from most frequent to least, although it varies a little). Assume that your key phrase will be made up mostly of common letters. For each letter pair, arrange it so that the more-common letter is on the top: S A D C E F T H I N ...

5

I mapped the symbols to $\{0,1,2\}$, so we get: 011 | 221 | 022 001 | 120 | 011 222 | 001 | 120 ----------------- 210 | 210 | 210 212 | 201 | 122 001 | 120 | 010 ----------------- 020 | 012 | 121 | 020 | ? 210 | 211 | I found this quite simple algorithm (which has two cases, either it is a row change [type2] or not [type1]): Type 1: * we (+1 ...

5

This is still an incomplete answer, but I'm getting closer. First, it's helpful to know that a cryptic clue has two components: A traditional "definition" of the solution, much like a "normal" crossword clue. Another way to create the same answer, which can be another simple definition, or more complex ways to build the answer, like word rebuses, "sounds ...

5

From my calculations... The way I approached it isn't elegant but I think my calculations are correct: Side note

5

This answer is correct, provided that they are standing in front of the doors (thanks for forcing the clarification @dcfyj). Take the following possibilities: The guard he talks to is a liar, and is in front of the castle door. He will respond "yes" to the question. This means that this is the correct door, because the truth-telling guard is not in front of ...

5

Let $R$ denote the radius of the circum-hypersphere and $r$ the radius of the in-hypersphere. We claim that To prove this, let $x_1,x_2,\cdots ,x_{n+1}$ be the $n+1$ points (these are really vectors in $\mathbb R^n$). They form a $n$ dimensional simplex. Now it's clear from symmetry that This proves our claim. Now of course,

4

20 cows. The 70 cows over the 24 days eat 1680 cow-nom units. The 30 cows over the 60 days eat 1800 cow-nom units. Based on the growth from 24 days to 60 days, we can work out that at 96 days, the cows would need to eat 1920 cow-nom units. Divide that by the 96 days and we get $20 cows$.

4

Those who study in the 6th class do not like both Maths and Biology is likely the correct interpretation because it leads to a unique solution. Just make a table as follows and apply the rules. Start with the most definitive statements such as the ones about Deepak and Farook.

4

One of the possible ways to solve such problem is to use integer programming. The idea is to transform your problem to a set of linear constraints on integer variables. For example, saying that a roll has a length of of 150m can be translated into a constraint that you cannot cut more than 150m out of the roll. So if the number of Marker1 you cut on this ...

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