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#341777 Checkerboard Challenge
Posted by bonanova on 02 March 2015  08:51 PM
#341748 A point in an equilateral triangle
Posted by bonanova on 28 February 2015  05:55 AM
And I suppose this result could be derived a la my last post.
I'm not sufficiently motivated.
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#341698 Squared rays
Posted by bonanova on 26 February 2015  01:55 AM
Yeah, I had to think that point through for a moment.
I saw it as having a sum that I didn't know how to evaluate.
So I added and subtracted the term that would make the sum zero.
That step simultaneously evaluated the sum and added 2 to the mix.
Your approach is neater, reasoning that the answer is not changed if you add the squared length of the n^{th} ray.
That's a nice insight.
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#341686 A simple integer function
Posted by bonanova on 24 February 2015  11:55 AM
Solution:
f (1) = 1? Can't be, since f (1) = f ( f (1) ) = 1, but it must be 3.
f (1) = 2? Could be, since f (2) would then be f ( f (1) ) = 3.
Must be. If f (1) were 3, then f (3) would also be 3. Nonincreasing.
Continue to use f ( f (n) ) = 3n:
 f (1) = 2
 f (2) = 3
 f (3) = 6 = f ( f (2) )
 f (4) =
 f (5) =
 f (6) = 9 = f ( f (3) )
 f (7) =
 f (8) =
 f (9) = 18 = f ( f (6) )
 f (10) =
We can fill in 4 and 5 because there are exactly two positive integers in the interval n = (3, 6), and exactly two other positive integers between f = (6, 9).
Values for n = 7 and 8 are then found using f ( f (n) ) = 3n:
 f (1) = 2
 f (2) = 3
 f (3) = 6 = f ( f (2) )
 f (4) = 7
 f (5) = 8
 f (6) = 9 = f ( f (3) )
 f (7) = 12 = f ( f (4) )
 f (8) = 15 = f ( f (5) )
 f (9) = 18 = f ( f (6) )
 f (10) =
Looking ahead to n = 12 we will have f = 21.
That constrains n = 10 and 11 to give f = 19 and 20.
Proceeding in this manner we obtain
 f (1) = 2
 f (2) = 3 = f ( f (1) )
 f (3) = 6 = f ( f (2) )
 f (4) = 7
 f (5) = 8
 f (6) = 9 = f ( f (3) )
 f (7) = 12 = f ( f (4) )
 f (8) = 15 = f ( f (5) )
 f (9) = 18 = f ( f (6) )
 f (10) = 19
 f (11) = 20
 f (12) = 21 = f ( f (7) )
 f (13) = 22
 f (14) = 23
 f (15) = 24 = f ( f (8) )
 f (16) = 25
 f (17) = 26
 f (18) = 27 = f ( f (9) )
 f (19) = 30 = f ( f (10) )
 f (20) = 33 = f ( f (11) )
 f (21) = 36 = f ( f (12) )
 f (22) = 39 = f ( f (13) )
 f (23) = 42 = f ( f (14) )
 f (24) = 45 = f ( f (15) )
 f (25) = 48 = f ( f (16) )
 f (26) = 51 = f ( f (17) )
 f (27) = 54 = f ( f (18) )
 f (28) = 55
 f (29) = 56
 f (30) = 57 = f ( f (19) )
We notice that when n is a power of 3, {1, 3, 9, 27, …}
f doubles it: f (3^{p}) = 2x3^{p}. e.g., f (27) = 54
We also notice than when n is twice a power of 3, {2, 6, 18, …}
f returns the next higher power of 3: f (2x3^{p}) = 3^{p+1}. e.g. f (2x9) = 27
To show that these relations hold for all p, we note first that they hold for p = 0 and 1 in the table.
The inductive step is completed by noting that
f (3^{p+1}) = f ( f (2x3^{p}) ) = 2x3^{p+1}
f (2x3^{p+1}) = f ( f ( 3^{p+1}) ) = 3^{p+2}
Now the intervening numbers (black numbers in the table) need to be accounted for.
There are 3^{p}  1 integers q in the interval q = (3^{p}, 2x3^{p}) and
there are 3^{p}  1 integers g in the interval g = (f (3^{p}) = 2x3^{p}, 3^{p}+1 = f(2x3^{p})).
So there are always just enough numbers to go around.
Thus, for q in [0, 3^{p}],
f (3^{p} + q) = 2x3^{p} + q and
f (2x3^{p} + q) = f ( f (3^{p} + q) ) = 3x(3^{p} + q).
Since 3^{6} = 729, we find that 2015 = 2x3^{6} + 557.
Thus f (2015) = 3x(3^{6} + 557) = 3,858.
 1
#341685 A simple integer function
Posted by bonanova on 24 February 2015  11:37 AM
#341593 The Price of Wine
Posted by bonanova on 16 February 2015  07:12 AM
Much more satisfying, as well! Good one.
 1
#341581 The Price of Wine
Posted by bonanova on 15 February 2015  06:21 AM
The three possible outcomes  lighter/equal/heavier tell you which type they are.
Example: the densities are 3 4 5.
3+3 < 4+5.
4+4 = 3+5.
5+5 > 3+4.
 1
#341521 kman's kasino
Posted by bonanova on 09 February 2015  09:25 PM
To celebrate the Grand Opening of his new casino, kman offered the first 1000 patrons the opportunity to win some pocket change by playing a game that carried the catchy phrase Flip while you're aHead. Each patron paid $100 to flip a coin multiple times, winning $100 for every H (heads) that appeared, and stopping (without penalty) at the first appearnace of a T (tails.)
What was kman's expected cost for this gesture of good will?
A coveted bonanova gold star for an answer from the The Book.
 1
#341493 The Knives and bottles table
Posted by bonanova on 03 February 2015  02:30 PM
Maybe just three knives would work.
The tips of the handles of the knives rest on the bottlenecks.
The blades interleave at the center of the triangle.
Each blade passes under the first, and above the second of the other two knives.
The blade tip of each knife rests on a blade of another knife, supporting the weight of the water.
So I wonder how to utilize a fourth knife..
 1
#341479 Polygons make a line
Posted by bonanova on 31 January 2015  01:44 AM
@Perhaps check it again
I thought of that distinction as well, and it led me back to the OP, where I found that the word "interior" is not mentioned.
It has been agreed that a line cannot always pass through the interior of polygons so described.
It has been clarified that BMAD asks something different.
Not all Denizens share English as their primary language, so discussions like this one are generally helpful if taken in that regard.
 1
#341390 The ngon eats out
Posted by bonanova on 15 January 2015  12:50 PM
n people party at a restaurant, sitting at the vertices of an ngonshaped dinner table. Their orders have been mixed up  in fact, none of them have received the correct entree. Show that the table may be rotated so that at least two people are sitting in front of the correct entree.
 1
#341346 I'm not a skier
Posted by bonanova on 01 January 2015  04:52 AM
 1
#341345 Empty a peg and marry a Princess (difficult)
Posted by bonanova on 01 January 2015  04:20 AM
I believe plasmid's line of analysis will lead to a solution.
To achieve a degree of closure this year, here is the 2^{nd} of the two proofs referenced in post 7.
It is much simpler than it first appears. Working an example is the best way to understand it.
It appeared in modified form on the Putnam Exam in 1993.
I found it in one of Peter Winckler's books (as a water and buckets problem) and adapted it to its present form.
The solution described below was found by Svante Janson of Uppsala University, Sweden.
It shows how the rings on one of the pegs can be reduced until it is zero.
bn
Label the pegs A, B, C with initially a, b and c rings, where 0 < a <= b <= c.
We describe a sequence of moves leading to a state where the minimum number
of rings on the three pegs is less than a. If this minimum is 0 we are done.
Otherwise we relabel and repeat.
Let b = qa + r, where 0 <= r < a, and q >= 1 is an integer.
Write q in binary form q = q_{0} + 2q_{1} + ... + 2^{n}q_{n} where each q_{i} is 0 or 1, and q_{n} = 1.
Do n+1 moves, numbered 0,...,n, as follows:
In move i we transfer rings
 from B to A if q_{i} = 1 or
 from C to A if q_{i} = 0.
Since we always move rings to A, its rings are doubled each time, so A has 2^{i}a rings before the i^{th} move.
Hence, the total rings moved from B equals qa, so at the end there remains (b  qa) = r < a rings on B.
The total rings moved from C is at most
{SUM _{i}^{n}_{=}^{}_{0}^{1} 2^{i}a} < 2^{n}a <= qa <= c,
so there will always be enough rings on C (and on B) to make these moves.
SJ
 1
#341238 Pawn Alice
Posted by bonanova on 07 December 2014  04:31 PM
 1
#341032 Upside Down Cake
Posted by bonanova on 12 November 2014  12:22 PM
This guy used Mathematica to simulate the problem.
For 185 degrees, it takes 4 cuts.
For 1 radian (an irrational number) it takes 84 cuts.
 1
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