14

Here is a straightforward escape route.


14

First: So, follow the instructions! (Thanks to El-Guest and hexomino for helping me fill in some of the final bits.)


12

I believe: My prefix is my abbreviation. My infix is. My suffix is seeing something. Together, I need someone like you. My prefix is a place. My infix is eating something. My suffix is suspicious by its presence. Together, you cannot let the one like me be nothing.


9

For each cycle, the number of prisoners executed is exactly A more mathematical approach:


9

The Solved Grid Solution Method A general observation: The first deduction we can make is in: Continuing: Next: Working in row 6 from the top: Continuing in this row: Some small deductions from here: Now let's try to place the 9's: Moving on: Let's tackle the upper-left box: Notice two important properties of these solutions: Some more deduction: ...


7



7

So the only unsolved ones from melfnt and Jeremy Dover, I think, are #1 and #5. #1 #5


7

Edit: Thanks to magma for pointing out some errors in my original calculation in the comments. Hopefully this new one makes more sense. I can present a semi-heuristic argument that the maximum probability achievable is And this can be achieved by choosing the curve to be, for example, Proof


7

This was fun, thanks for posting it - hopefully my answer is correct! Explanation:


6

A general [no-computers] solution: First of all: To summarize: Let's now consider iterating this. The following diagram may help to illustrate what happens: Let's now apply this to the present case.


6

Doesn't look like anyone has answered the "no computers" part yet, so I will do that. First, a simple observation: That also means: This has an important consequence: I'm going to call the most significant digit from such a multiplication the "carry" from here on, and the least significant digit "LSD". Single digit ...


6

Wife's busy, so maybe I can sneak this in: #2: #3: #6: #8:


4

Partial answer (3/8) (solved) (solved)


2

$533.(3)$ Let's do it as follows: Last span will be $x$ miles, and final answer is $1000-x$ bananas delivered. Next to last span will be $y$ miles, we would do best if each ride starts with 1000, i.e. $2000 -3y = 1000$ should hold to get to 1000 left for the last span. Hence $y = \frac{1000}{3}$. Now to get to 2000 from 3000 we will need 3 trips on path of ...


2

A quick and dirty approach, slightly different to the ones given. Again, [no-computers].


2

The three printouts have cent values of Why?


2

Let's first see in how many ways the numbers can be placed with all neighbours coprime. Now for the probability:


2

You misinterpreted the puzzle. You shouldn't place a number in the last slot; you should move them around so that the numbers are in order, i.e. from the shown position you need to swap the 11 and 10. It's a type of sliding puzzle, more than a century old, known as the 15-puzzle. Half of the starting positions of this puzzle are unsolvable, as is this ...


1

I think the questions and answers are essentially correct as written. I don't speak German, so I can't say whether or not the German words are used correctly, but if I make some reasonable assumptions about what everything means, then I get the marked answers. (By the way, you seem to have a translation error in the last one; the German says "gegen den ...


1

Through the centre O of the great circle with r=5 units we draw a straight line AB which passes through the centre O. We obtain from the centre on both sides lengths equal to 3.4 units. After that, we draw a straight line CD perpendicular to the straight line AB both passing through the centre O. We obtain lengths from the centre OC=2.95 units and OD=2.95 ...


1

The answer is because


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