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Posts posted by jamieg


It's the German Tank Problem:
N~=m1+m/k, where N is total taxis, m is highest number, and k is number of taxis spotted.
N~=901+90/5=107 taxis.Based on probability mass function, approach gives expected value of
N=(m1)*(k1)/(k2)
N=89*4/3=119.http://en.wikipedia.org/wiki/German_tank_problem#Example

The next palindrome is 26062, which is 110 miles later. Over two hours of driving, this was 55 mph.

Is the number of skittles in the image correct? You don't otherwise provide how many are being poured. Or do you want a general solution for arbitrary N?

To make it solvable, change the random scale into an alternating scale.
With an alternating scale it becomes simple; after your third weighing, use either other scale once. If the scale gives a different result than before, it's alternating.

I don't see how this is possible. Theoretically the random scale has a small possibility of replicating either of the constant scales indefinitely just by random chance. If it randomly gives the correct answer the first ten times, any observation on it will be the same as if it were the truth telling scale; same with if it's a hundred or infinite times.
You can determine the identity of a single scale in 3 weights. Weigh any one scale with a single coin on one side and no coins on the other side. The correct one will always weigh it the correct way, the false one will always weigh it the wrong way, and the random one can be either. Whichever one is unique to the other two can be determined (e.g. 2 correct answers and 1 incorrect means the incorrect is definitely the bad scale).

Trivial case excluded?
(Rectangular box of sides nxnxn, filled with a single cube of sides nxnxn, no two cubes have the same length)

There's an even simpler observation that immediately gives the correct answer.
I assumed there is,
2
^{9}1, but I'm not sure right now why that is.

First group it by number of digits (so group 1 will be 1 digit, group 2 will be 2 digits, etc). Then each group will be the functional equivalent of asking, "what are the possible combinatorics for n numbers?" where n is the group number. The reason for this is that any combination of (nonrepeating) digits can be ordered from lowest to greatest, and the combination function (nCr) basically takes this into account (as opposed to a permutation, in which the order of items creates unique possibilities). So the final solution will be
SUM (from n=1 to n=9) of 9Cn
=511

Each time you make a random cut, you'll intersect all of the previous cuts. Some of the pieces may be very small, but there will still be a piece, as the probability of intersecting at a previous intersection is 0.
The number of pieces, therefore, is one more than the sum of 1, 2, 3, 4, ..., N (since zero cuts gives you one piece), which comes out to N (N+1) / 2 + 1
The cuts are random, not necessarily maximizing number of slices. Otherwise you'd be correct.

Simplifying all equations without any solutions to make things easier:
n^{n1}/2
sqrt(n)1
sqrt(n)+1
n^{2}
(3n)!n
(3(n!))!+n
2srqt(n)+1
n*1^{n} (=n if n<infinity)
2+n
2n+1
2n+(n1)!
2n+n (=3n, n>0, or =n, n<0)
2n*.n+.n (=.n*(2n+1))
2n
(n+1)/.n
2n(sqrt(n)1)!
2n+1
2n^{n}+n
(n1)!n
2n!1
n*sqrt((n.n)/.n)
n(1+2/.n)
sqrt(.n^{}^{n})*nsqrt(n)
(nn/nn)!
n^{2}

Nicely done on mine, googon97. Bonanova, I've solved yours:
Next:f(x,y)=(x+1)
^{2}  (y1)^{2} + x(y2)x\y 1 2 3 4 5 6
+
0  4 10 28 82 244 730
1  1 7 25 79 241 727
2  7 13 31 85 247 733
3  5 1 19 73 235 721
4  19 25 43 97 259 745
5 29 23 5 49 211 697
6  67 73 91 145 307 793 
I'll add my own now:
x\y 1 2 3 4 5 6
0 2 2 2 2 2 2
1 1 0 1 2 1 0
2 0 2 0 2 0 2
3 1 0 1 2 1 0
4 2 2 2 2 2 2
5 1 0 1 2 1 0
6 0 2 0 2 0 2

Very cool puzzle, I'd love to see more (some harder, but some at this level as well).
The first row really gives the first term awayonly y^x could produce 1s across the board at x=0 and nontrivial values after. Then the fact that it's and odd function (or whatever you'd call a multivariable function with rotational symmetry over y=x) gives
f(x,y)=y^xx^y

Start by assuming the final rectangle must have integer values. The total area of the mentioned rectangles is 1056 (source). The possible rectangles with integer values with area 1056 will be the pairings of the lowest factor of 1056 with its highest factor, or its secondlowest with its secondhighest, or thirdlowest/thirdhighest, etc:
(1, 1056); (2, 528); (3, 352); (4, 264); (6, 176); (8, 132); (11, 96); (12, 88); (16, 66); (22, 48); (24, 44); (32, 33). (source)
We can rule out any with width<18 because it couldn't possibly hold our largest square, leaving us with just the final 3:
(22, 48); (24, 44); (32, 33).
We would have to rule out any with a length less than 15+18=33, but all three pass this test.
We then have to assume for just a second that the 14piece will be found somewhere adjacent to the 15piece (if the 18 piece is in the bottom left and the 15 piece in the bottom right, the most efficient place for the 14piece in order to keep a small border would be above the 15 piece). This means the width must be at least 29, ruling out (22,48) and (24,44), leaving us with just (32,33).
Edited: (22,48) is possible by putting the three largest pieces lengthwise as 15+18+14=47, but this would obviously lead to problems when trying to fill up to hit the 48th unit lengthwise as we have only one 1piece.
Now that we have the shape of the final rectangle, trial and error (with some common sense to speed things up)gives us the final shape. Click here for final.

Maximum is 48 drops with 2 bulbs:
drop at floor 48. If bulb breaks, increments of 1 from floor 1 to floor 47 (max 48 if breaks at floor 47). If bulb survives first drop, go to floor 48+47=95 (2nd drop). If break, increments of 1 from 49 to 94 (max 48 if breaks at floor 93). If bulb survives at 95, go to floor 48+47+46=141. If breaks at 141 (third drop), incremental from 96 to 140 (max 48 if breaks at floor 140).
Note that the pattern will get you to 48+47+46+45+...+3+2+1, which is overkill (sum=1035), but starting at 47 won't get you to 1000 floors. So 48 is our magic number.
As far as extrapolating for future steps, my instinct would be to do a similar procedure for n=3, where, if the bulb breaks (eg) at 48, then you go to floor 10, then if break incremental; if not, floor 10+9=19, etc etc, which will have a max of 10 within the step, for a max of 11; however, 11 isn't our maximum because we know it can take 48 drops to get to the top floors. So there's some other way that takes these steps to make for even distribution regardless of the floor, e.g. floor 14^2, then floor 14^2+13^2, then 14^2+13^2+12^2, etc, and then a pattern within that following the n=2 system (N+(N1)+(N2)...+1). But I'll let someone else do the actual optimization and give all the glory to him/her/them. 
Define the ant's rate of travel as its constant speed plus the speed of the band where it stands.
Constant speed = 0.1
Speed of band is the ratio of the ant's current location (x) to the location of the truck times 10mph (assuming the band's stretch is consistent from start to finish). The current location of the truck is 10t, so the speed of the band is x/t.
dx/dt=0.1+x/t
Solving the ODE gives x=t(c+0.1*ln(t))
Given x(.01)=0, (from: truck going 10mph, hits 0.1m at 0.01h)
0=.01*(c+.1ln(.01)), c=0.1ln(0.01)
Finding intersection:
0=t(0.1ln(0.01)+0.1*ln(t))10t
gives, according to wolfram alpha,
x~~0.0003205380.000248259 i
Which, unless something went wrong, means poor Andy will never reach that truck.
Changing times
in New Logic/Math Puzzles
Posted
Can you clarify  are you looking for highest number of coins, or the highest total value of coins? Are you including coins out of circulation?