29
The given sum is what we would get if we were to compute the definite
But we can also compute this directly:
Now by amazing coincidence $9.3$ squared happens to evaluate to
which is good enough to pass for $86.5$. One more lucky coincidence: $9.3$ is easily multiplied by $\frac 23$. So we need to do $6.2$ times $86.5$ which yields
The $.3$ cancels ...
17
It is not possible to move all 16 bars.
Consider the total distance that each person moves, in each direction, carrying each possible number of bars. For example $R_2^+$ is the total distance that Rod moves forward while carrying 2 bars. We can set up a system of equations. Both Rod and Lia must move a net distance of $1$ mile forwards:
$$
R_0^+ - R_0^- + ...
13
One solution meeting your requirements:
11
Yet another one, with all 16 words different: (I admit that this is a computer-assisted answer, in the sense that I wrote a program to extract the list of 4-letter heteropalindromes from YAWL. I constructed the square manually from that list.)
For side information: there are only 16 unique such squares containing 16 unique words (not counting rotations and ...
10
You can turn Life into Shit in 6 steps:
or
or
This is the optimal solution.
Why 6 steps?
My thinking solution:
9
This approach brings 14 bars and leaves them 20 minutes before the train arrives.
This lets them bring a 15th bar and catch the train with less than 5 minutes to spare.
8
From OEIS: A000448
For a circle centered at the origin, this gives these lattice points:
For a total of
7
I'll take a shot at it
Arrived at as follows
First
Then
So
But
So
7
Deuosvi had already posted an answer, but I couldn't stop myself when I found another one:
5
Since black has many checks available, we can
This limits our options to four possible moves. Nf7+ seems particularly promising, so
Black has only one move that doesn't immediately end in a smothered checkmate at Nf6, so we check the checks (heh) after that move to find
after which we can somewhat incredibly finish with either
or
3
Here is the finished tiling:
To get started:
Next step:
Looking at the left pear:
With those chokepoints:
For the next step I took a guess:
From there, I just looked at the pieces I had left and found something that worked.
3
A little experimentation with Euclids formula, demonstrates that there are only
primitive Pythagorean triples with largest element less than or equal to 80. If, from these, we choose the sets
and note that 65=5*13, then we can produce the triples
Using them, plus their reflections about the lines y=x, y=-x, y=0, and x=0, plus the four extra points sitting ...
2
An alternative method to Kate's to get the same amount of bars.
1
Sure, a 1×4√5 rectangle and a 1×√5 rectangle.
1
The answer is:
Method:
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