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
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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